Chemical_Reactions_and_Stoichiometry_230517_094230_(1).pdf

this topic shows stoichiometry

1. CHY 30 Mass Relationships in Chemical Reactions 1 of 9
MASS RELATIONSHIPS IN CHEMICAL REACTIONS

OBJECTIVES
This unit introduces the quantitative relationships of reactants and products in a chemical reaction
and it aims that the students will be able to
1. understand the concept of a mole and molar mass;
2. determine the atomic mass of an element and formula mass of a compound;
3. calculate for the quantities of reactant or product from a given quantity of the other;
4. determine the limiting and excess reactants; and
5. determine the percent yield of a chemical reaction.

LAWS OF CHEMICAL COMBINATION
1. Law of Conservation of Mass (Antoine Lavoisier, 1774)
During a chemical reaction, the total mass before reaction is equal to the total mass after
reaction. Conservation means something can neither be created nor destroyed. Here, it applies
to matter.

2. Law of Definite Proportions (Joseph Proust, 1799)
The relative kinds and numbers of atoms are constant for a given compound. A compound
always has the same formula.

3.
Law of Multiple Proportions (John Dalton, 1803-1807)

If two elements A and B combine to form more than one compound, then the mass of B which
combines with the mass of A is a ratio of small whole numbers.

ATOMIC AND MOLAR MASS
Atomic mass scale - a relative scale originally based on H which is assigned a mass of 1.
Atoms are too small to weigh directly. For example, one carbon atom has a mass of 1.99 x 10
-23
g
– too inconvenient! We need a more convenient unit for mass.

Atomic mass units are convenient units to use when dealing with extremely small masses of
individual atoms.

1 amu = 1.66054 x 10
–24
g and 1 g = 6.02214 x 10
23
amu

Carbon-12 was chosen and given a mass value of 12 amu. By definition, the mass of
12
C is exactly
12 amu. Mass of all other atoms measured relative to mass of carbon-12.

1 amu = 12
1 the mass of carbon-12

Atomic Mass - the mass of atoms of elements expressed in atomic mass units.

1 g = 6.022 x 10
23
amu

The mass of carbon on the periodic table is 12.01 amu, NOT 12.00 amu - WHY?!
Atomic masses reported on the periodic table are weighted averages of all naturally occurring
isotopes for each element.

Isotopes - two or more atoms (nuclide) having the same atomic number but different mass number
(e.g. Ne
20
10 , Ne
21
10 , and Ne
22
10 ).

Average atomic mass/atomic weight - is the average of the isotopic masses, weighed
according to the naturally occurring abundances of the isotopes of an element.
Atomic Weight = n
i
i=1
(fractional abundance x mass of isotope)
Note: fractional abundance = percent abundance / 100

Example: Naturally occurring chlorine is 75.53 percent
35
Cl, which has an atomic mass of 34.969
amu, and 24.47 percent
37
Cl, which has an atomic mass of 36.966 amu. Calculate the average
atomic mass (that is, the atomic weight) of chlorine.

Solution: Average atomic mass = (0.7553) (34.969 amu) + (0.2447) (36.966 amu)
= 35.5 amu
EXERCISE 1
1. The three stable isotopes of neon: Ne
20
10 , Ne
21
10 , and Ne
22
10 , have respective abundances of
90.51%, 0.27% and 9.22%. Obtain the average atomic mass of neon.
2. If 98.89% of carbon exists as carbon-12, which has a mass of 12.000000 amu, while 1.11%
exists as carbon-13, which has a mass of 13.00335 amu, calculate the average atomic mass for
carbon.

3. Two naturally occurring isotopes of boron, boron-10 and boron-11, have the masses of
10.012937 amu and 11.009305 amu, respectively. Calculate the percent natural abundance of
each isotope given the atomic mass of boron as 10.81 amu.

Molecule - an assembly of two or more atoms (nonmetals) tightly bound together. It is a group of
bonded atoms that actually exist and can be identified as a distinct entity.

Molecular compound - compounds that are composed of molecules and contain more than one
type of atom (e.g. H2O, CCl4). Most molecular substances contain only nonmetals.

Ionic compounds - combinations of metals and nonmetals; a compound made up of positive
(cation) and negative (anions) ions joined together by electrostatic forces of attraction.

Chemical Formula - a combination of symbols used to represent a compound.
It indicates:
a. the elements present
b. the relative numbers of atoms of each element in the compound - indicated by a subscript
following the element’s symbol. If there is no subscript, only one atom of that element is in the
compound.

2. CHY 30 Mass Relationships in Chemical Reactions 2 of 9
Molecular Formula - chemical formulas that indicate the actual numbers and types of atoms in a
molecule.

Formula weight (FW) - is the sum of atomic weights for the atoms shown in the chemical
formula.

Molecular weight (MW) - is the sum of the atomic weights of the atoms in a molecule as shown
in the molecular formula.

Example: Calculate the formula weight or molecular weight of the following:
1. H2O
MW = (no. of H atoms from chemical formula x atomic weight of H from periodic table) +
(no. of O atoms from chemical formula x atomic weight of O from periodic table)
= (2 x 1.01 amu) + (1 x 16.00 amu)
= 18.02 amu

2. Ca(OH)2
FW = (1 x 40.08 amu) + (2 x 16.00 amu) + (2 x 1.01 amu) = 74.10 amu

3. BaCl2 ∙ 2H2O (barium chloride dihydrate)
FW = (1 x 137.33 amu) + (2 x 35.45 amu) + (4 x 1.01 amu) + (2 x 16.00 amu)
= 244.27 amu

EXERCISE 2
Calculate the formula/molecular weight of the following:
1. HCl 3. CH3CH(NH2)COOH (alanine)
2. C6H12O6 4. CuSO4 ∙ 5H2O

Percentage composition from formulas - the percentage by mass contributed by each element
in the substance.
% �=
(??????�.�?????? � ??????����)(??????� �?????? �)
??????� �?????? ��������
?????? ���

Steps in determining percentage composition:
1. Determine the mass of each individual element in the compound. Refer to the periodic table.
2. Add up all the masses of each element to get the total mass of compound.
3. Divide the mass of each individual element with the total mass of compound.

Example: Calculate the percentage of each element in H2O.
Solution: FW of H2O = (2 x 1.01 amu) + (1 x 16.00 amu)
= 18.02 amu






EXERCISE 3
Calculate the percentage composition of each element in:
1. CCl3Br 2. Fe4[Fe(CN)6]3 3. PCl5 4. C12H22O11

Answer (#2)
FW of Fe4[Fe(CN)6]3 = 7 (55.85) + 18 (12.01) + 18 (14.01) = 859.31

% Fe = [ 7 (55.85) / 859.31 ] x 100 = 45.46
% C = [ 18 (12.01) / 859.31 ] x 100 = 25.16
% N = [ 18 (14.01) / 859.31 ] x 100 = 29.35


MOLAR MASS AND AVOGADRO’S NUMBER
Mole (n) - is the amount of matter that contains as many particles (atoms, molecules, or ions) as
the number of atoms contained in exactly 12 g of
12
C.

Avogadro’s Number (NA) = 6.022 x 10
23
entities

that is, 1 mol of
12
C  6.022 x 10
23
C atoms
1 mol of HC2H3O2  6.022 x 10
23
HC2H3O2 molecules
1 mol of OH
-
 6.022 x 10
23
OH
-
ions
1 mol of MgCl2  6.022 x 10
23
MgCl2 formula units

Molar mass - the mass in grams of 1 mole of substance (g/mol).
Molar masses for elements are the same as their atomic weight found on the periodic table. (1 mol
= 6.02 x 10
23
is the amount of atoms of any element that has a mass in grams equal to the mass
of one atom in amu).

The molar mass of a molecule is the sum of the molar masses of the atoms; thus, has same value
as its molecular weight. The formula weight is also numerically equal to the molar mass.
The atomic masses reported for each element in the periodic table gives the atomic weight in
amu and the molar mass in g/mol.

One H2O molecule weighs 18.0 amu → 1 mol H2O weighs 18.0 g
One NO3
-
ion weighs 62.0 amu → 1 mol NO3
-
weighs 62.0 g
One NaCl unit weighs 58.5 amu → 1 mol NaCl weighs 58.5 g

Example: What is the molar mass of Mg3(PO4)2?
Answer: Molar mass of Mg3(PO4)2 = (3 x 24.31 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol)
= 262.87 g/mol

EXERCISE 4
1. Determine the atomic/molar mass of the following:
a. Al c. Cu2O e. Al2(SO4)3
b. CuO d. CuCl2

3. CHY 30 Mass Relationships in Chemical Reactions 3 of 9
USEFUL CONVERSIONS

Mass = Moles x Molar mass




No. of particles = Moles x NA
particles = mol x particle/mol
Examples:
1. How many grams are there in 3.80 moles of H2O?
Answer:



2. How many molecules are there in 3.80 moles of H2O?
Answer:



3. How many molecules of H2O are there in 7.4 g of H2O?
Answer:

4. How many grams H2O are contained 8.956 x 10
24
molecules

of H2O?
Answer:



EXERCISE 5
1. a. How many moles of glucose (C6H12O6) are there in 538 g of the substance?
c. How many glucose molecules are there in 5.23 g of glucose?
2. a. How many moles of Ne are there in 0.500 g Ne?
b. How many Ne atoms are there in 0.500 g Ne?
3. How many moles of O2 contain 5.23 x 10
24
molecules of oxygen?
4. What is the mass of 0.055-mol water whose molar mass is 18.02 g/mol?

EMPIRICAL FORMULAS FROM ANALYSES
Empirical formula (EF) - simplest whole-number ratio of atoms in a substance; may or may not
be the same as the molecular formula.

Molecular formula (MF) - represents the molecule of a substance and shows the actual number
of atoms in a molecule.
MF = (EF) n where n = EW
MW

Where MW is the molecular weight and EW is the empirical weight (i.e., the sum of atomic weights
in the empirical formula)

Table 1. Examples of Empirical and Molecular Formulas
Empirical Formula Molecular Formula n
CH3 C2H6 2
CH2O C6H12O6 6
HO H2O2 2
CH C6H6 6
CH2O CH2O 1

Note that substances with similar empirical formulas have the same percentage composition.




Figure 1. Diagram for empirical formula calculation

Example: Dibutyl succinate is an insect repellant used as agents for household ants and roaches.
Its composition is 62.59% C, 9.63% H, and 27.79% O. Its experimentally determined molecular
weight is 230 amu. What are the empirical and molecular formulas of dibutyl succinate?

Solution:
Assume 100 g sample: C = 62.59 g
H = 9.63 g
O = 27.79 g

Therefore, the empirical formula of dibutyl succinate is C6H11O2
and the empirical weight is EW = (6 x 12.01) + (11 x 1.01) + (2 x 16.00) = 115.17
massMolar
Mass
Moles= molg
g
mol
/
= mol
g
molxg=

4. CHY 30 Mass Relationships in Chemical Reactions 4 of 9
The molecular formula is









EXERCISE 6
1. Ascorbic acid (Vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the
empirical formula of ascorbic acid?
2. Sorbitol, used as sweetener in some sugar-free foods, has a molecular weight of 182 amu and
a mass composition as follows: 39.56 % C, 7.74 % H, and 52.7% O. What are the empirical
and molecular formulas of sorbitol?

ESTABLISHING EMPIRICAL FORMULA FROM EXPERIMENTAL DATA
Combustion Analysis OHCOOOHC
zyx 222 +→+

All C atoms in the compound → C atoms in CO2
All H atoms in the compound → H atoms in H2O
All O atoms in the compound and the O2 gas consumed → O atoms in CO2 and H2O

Step in determining the empirical formula using data from combustion analysis:
1. Convert: grams CO2 → grams C
grams H2O → grams H
Calculate: grams O = grams sample – (grams C + grams H)
2. Convert: grams C → moles C
Grams H → moles H
grams O → moles O
3. Calculate the mole ratio and determine the empirical formula.
4. From the empirical formula, calculate the formula weight and calculate n from the formula
weight and the molecular weight.

Example: Caproic acid, the substance responsible for the aroma of dirty gym socks and running
shoes, contains C, H, and O only. Combustion of 0.450 g sample of caproic acid gives 0.418 g H2O
and 1.023 g CO2. What is the empirical formula of caproic acid? If the molecular weight of caproic
acid is 116.2, what is the molecular formula?
Solution:
Calculate the amount (in grams) of the elements C, H, and O as follows:








EXERCISE 7
1. Consider a sample of isopropyl alcohol, a substance sold as rubbing alcohol. The compound
is known to contain only C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces
0.561 g CO2 and 0.306 g H2O. Determine the empirical formula of isopropyl alcohol. If the
molecular weight of isopropyl alcohol is 60.11, what is the molecular formula?
2. A sample of the black mineral hematite, an oxide of iron found in many iron ores, contains
34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?
3. The bacterial fermentation of grain to produce ethanol forms a gas with a percent
composition of 27.29% C and 72.71% O. What is the empirical formula for this gas?
4. Polymers are large molecules composed of simple units repeated many times. Thus, they
often have relatively simple empirical formulas. Calculate the empirical formulas of the
following polymers:
(a) Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O (d) polystyrene; 92.3% C, 7.7% H
(b) Saran; 24.8% C, 2.0% H, 73.1% Cl (e) polyethylene; 86% C, 14% H
(c) Orlon; 67.9% C, 5.70% H, 26.4% N

5. CHY 30 Mass Relationships in Chemical Reactions 5 of 9
CHEMICAL REACTIONS AND EQUATIONS
Stoichiometry - from the Greek word stoicheion (element) and metron (measure). It deals with
the quantitative relationship of chemical formulas and chemical reactions. Alternatively, it deals with
mass relationships between the reactants and products in a chemical reaction.

Chemical reaction - a process by which a set of substances is transformed through a chemical
change into a new set of substances.
Physical manifestations of a chemical reaction:
1. color change
2. formation of a solid (precipitate) within a clear solution
3. evolution of gas
4. evolution or absorption of energy

Chemical equations - give a description of a chemical reaction.
There are two parts to any equation:
1. Reactants (written to the left of the arrow)
2. Products (written to the right of the arrow)

Generally, a chemical reaction is represented by this equation:
Reactants → Products )()()()( zyxw dDcCbBaA +→+

where: A, B, C, and D are chemical species
a, b, c, and d are stoichiometric coefficients
w, x, y, and z are physical state of the substance in a reaction

There are two sets of numbers in a chemical equation:

1. Numbers in front of the chemical formulas (stoichiometric coefficients) - give the ratio in
which reactants and products exist.
2. Numbers in the formulas (subscripts) - give the ratio in which the atoms are found in the
molecule.

Illustration: Hydrogen combines with oxygen to form water.

Atomic/molecular level:
2 H2 molecules + 1 O2 molecule → 2 H2O molecules
Mole level: 2 H2 (g) + O2 (g) → 2 H2O (l)
Mass level: 2 (2.02 g/mol) + 1 (32.00g/mol) → 2 (18.02g/mol)

The subscript 2 in H2O means there are two H atoms for each molecule of water.

The coefficient 2 in 2 H2O means that there are two water molecules present. In 2 H2O there are
a total of four hydrogen atoms present (two from each water molecule).

From the balanced chemical equation, it can be deduced that 2 mol H2 = 1 mol O2 = 2 mol H2O and
are therefore stoichiometrically equivalent quantities.

Types of Chemical Reactions
1. Combination 4. Double Displacement
2. Decomposition 5. Combustion
3. Single Displacement 6. Oxidation-Reduction (Redox)

Stoichiometric Calculations


Figure 2. Diagram for stoichiometric calculations

Example: Given the reaction: )(2)()(2
222 lOHgOgH →+
(a) How many moles of H2O can be produced from 1.57 moles O2?
Solution:







(b) How many grams of O2 is needed to completely react with 4.2 moles of H2?
Solution:

6. CHY 30 Mass Relationships in Chemical Reactions 6 of 9

(c) How many grams of H2O will be produced from 6.15 g of H2?

Solution:











EXERCISE 8
1. Consider the following reaction: )(2)()(2)(
2224 gOHgCOgOgCH +→+
a. How many moles of O2 are needed to completely react with 2.25 moles of CH4.
b. How many moles of CO2 are formed when 5.25 moles of O2 completely react?
2. Given the unbalanced equation: )()()(
322 sOAlgOsAl →+
a. How many moles of Al and O2 are needed to produce 10.0 mol Al2O3?
b. How many moles of O2 are required to produce 158.00 g Al2O3?
c. How many grams of Al and O2 are needed to yield 86.52 g Al2O3?
3. What mass of sodium hydroxide, NaOH, would be required to produce 16.0 g of the antacid
milk of magnesia [magnesium hydroxide, Mg(OH)2] by the following reaction?
MgCl2(aq) + 2NaOH(aq) ⟶ Mg(OH)2(s) + 2NaCl(aq)
4. The human body needs at least 1.03 x 10
-2
mol O2 every minute. If all of this oxygen is used
for the cellular respiration reaction that breaks down glucose, how many grams of glucose does
the human body consume each minute?
C6H12O6(s) + 6 O2(g) ⟶ 6 CO2(g) + 6 H2O(l)

Limiting reagent (LR) - the reactant that determines the extent at which a chemical reaction
proceeds, the amount of products consumed, and the amount of products formed.

Excess reagent (ER) - the reactant presents in quantities greater than what is needed to react
with the quantity of the limiting reagent present.

Guidelines in the calculation of problems involving limiting reagent:
1. Convert given amount of reactant to moles.
2. Divide mole of each reactant by their corresponding coefficients from balanced chemical
equation. The one with smallest result is the limiting reagent.
3. Calculate the required quantity based on the given amount of limiting reagent.

Example: How many grams of NO can be produced in the reaction of 5.00 g NH3 and 7.50 g O2? )(6)(4)(5)(4
223 lOHgNOgOgNH +→+


Solution:















EXERCISE 9
1. Consider the reaction: )(2)(3)(2
32 sAlClgClsAl →+
A mixture of 1.5 grams Al and 3.0 grams Cl2 are allowed to react.
a. What is the limiting reagent?
b. How many grams of AlCl3 are formed?
2. Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of
turbine blades in jet engines. It is prepared according to the following equation:
3 Si(s) + 2 N2(g) ⟶ Si3N4(s)
Which is the limiting reactant when 2.00 g of Si and 1.50 g of N2 react?

3. 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Here is
the balanced equation for the reaction:
Al2S3 + 6 H2O ⟶ 2 Al(OH)3 + 3H2S
(a) Which is the limiting reagent?
(b) What is the maximum mass of H2S which can be formed from these reagents?
(c) How much excess reagent remains after the reaction is complete?

7. CHY 30 Mass Relationships in Chemical Reactions 7 of 9
Theoretical yield (TY) - the calculated amount of products expected from given quantities of
reactants in a chemical reaction.

Actual yield (AY) - the amount of product actually produced.
The actual yield may not be the same as the theoretical yield for some reasons:
• Some products may be lost during purification steps → decreases AY
• Formation of by-products during side reactions → increases AY
• Products formed may be wet → increases AY
• Products may be contaminated with excess reagents → increases AY

PERCENTAGE YIELD 100
Yield lTheoretica
Yield Actual
Yield % x=


Example: Titanium is prepared by the reduction of titanium (IV) chloride with molten magnesium
between 950C and 1150C. In a certain operation, 3.54 x 10
4
kg TiCl4 is reacted with 1.13 x 10
4
kg
Mg according to the following reaction: )(2)()(2)(
24 lMgClsTilMggTiCl +→+

a. Calculate the theoretical yield of Ti in kg.
b. What is the percentage yield if 7.91 x 10
3
kg Ti is actually produced?



















EXERCISE 10
1. The reaction of 15.0 g C4H9OH, 22.4 g NaBr, and 32.7 g H2SO4 yields 17.1 g C4H9Br in the
reaction: OHNaHSOBrHCSOHNaBrOHHC
24944294 ++→++
a. Determine the limiting reagent and excess reagents.
b. Calculate the theoretical yield and the percentage yield.

2. A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of silver
nitrate, causing the following reaction to occur: )()()(2)(2)(
233 aqNOZnsAgaqAgNOsZn +→+

a. Which is the limiting reactant?
b. How many grams of Ag will form?
c. How many grams of Zn(NO3)2 will form?
d. How many grams of the excess reagent will be left at the end of the reaction?

3. Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was
obtained according to the equation: CuSO4(aq) + Zn(s) ⟶ Cu(s) + ZnSO4(aq)
What is the percent yield?


ADDITIONAL EXERCISES
1. Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?

2. What is the empirical formula of a compound which contains 60.0% O and 40.0% S by mass?

3. Determine the empirical and molecular formula of the following compounds:
a. Ibuprofen, an analgesic, which contains 75.69% C, 8.80% H, and 15.51% O by mass and
has a molar mass of 206 g/mol.
b. Monosodium glutamate (MSG), a flavor enhancer, which contains 35.51% C, 4.77% H,
37.85% O, 8.29% N, and 13.60% Na and has a molar mass of 169 g/mol.

4. There are four naturally occurring isotopes of chromium. Their masses and percent natural
abundances are 49.9461 u, 4.35%; 51.9405 u, 83.79%; 52.9407 u, 9.50%, and 53.9389 u,
2.36%. Calculate the atomic mass of chromium.

5. Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250 mg sample of nicotine
was combusted producing 14.242 mg CO2 and 4.083 mg H2O. What is the empirical formula of
nicotine? If the substance has a molar mass of 160 ± 5 g, what is the molecular formula?

6. Vanillin, the dominant flavoring in vanilla, contains three elements: C, H, and O. When 1.05 g
of this substance is completely combusted, 2.43 g CO2 and 0.50 g H2O are produced. What is
the empirical formula of vanillin?

8. CHY 30 Mass Relationships in Chemical Reactions 8 of 9
7. An element X forms a compound with carbon for which the formula is XC2. If 37.48% of the
compound is carbon, what is the atomic weight of X?

8. Mercury has a density of 13.53 g/cm
3
. How many moles of mercury are in a 25-cm column
with a radius of 2.54 cm?

9. Consider stearic acid, C18H36O2

(284.4 g/mol).
a. The mass in grams of 0.450 mol of stearic acid is ____________.
b. The number of moles of H atoms in 2.4 x 10
24
molecules of stearic acid is ____________.
c. The mass percentage of oxygen in stearic acid is ____________.
d. The total number of atoms in 0.150 mol of stearic acid is ____________.

10. Consider the following: Hydrogen cyanide, HCN (27.0 g/mol), is prepared from ammonia, NH3
(17.0 g/mol), oxygen, O2 (32.0 g/mol), and natural gas, CH4 (16.0 g/mol), by the following
process: )(6)(2)(2)(3)(2
2423 gOHgHCNgCHgOgNH +→++ .
a. The amount in grams of HCN that can be produced from the reaction of 1.00 mol of NH3
with sufficient O2 and CH4 is ____________.
b. The amount in grams of O2 that will be needed to react completely with 20.0 g CH4 is
____________.
c. The amount in grams of H2O produced for every 5.00 g of HCN formed is ____________.
d. If 1.00 mol each of NH3, O2, and CH4 are mixed, which is the limiting reactant?
e. The amount in moles of HCN that can be produced in (d) is ____________.
f. The amount in grams of CH4 needed to produce 1.00 mol HCN if the reaction has a yield
of 82.0% is ____________.

11. Determine the molar mass of the following compounds:
a. C2H5OH c. AlCl3
b. Al2(SO4)3 d. N2O4

12. Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component
elements: )(3)(22
23 gNsNaNaN +→
a. How many moles of nitrogen gas are produced by the decomposition of 1.50 mol NaN3?
b. How many grams of NaN3 are required to form 5.00 g nitrogen gas?
c. How many grams of NaN3 are required to produce 100 L nitrogen gas if the gas has a
density of 1.25 g/L?

13. How many grams of carbon dioxide can be formed when a mixture of 4.95 g ethylene (C2H4)
and 3.25 g O2 is ignited, assuming complete combustion to form carbon dioxide and water.

14. One of the steps in the commercial process for converting ammonia to nitric acid involves
the conversion of NH3 to NO: 4 NH3 + 5 O2 → 4 NO + 6 H2O
In a certain experiment, 2.50 g NH3 reacts with 2.85 g O2.
a. Which is the limiting reactant?
b. How many grams of NO are formed?
c. How much of the excess reagent remains after the limiting reactant is completely
consumed?
15. A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of silver
nitrate, which deposited silver and formed
Zn + 2 AgNO3 → 2 Ag + ZnNO3
a. Which is the limiting reactant?
b. How many grams of silver is expected to form?
c. How many grams of zinc nitrate is expected to form?
d. How many grams of the excess reactant is left after the reaction?
e. If the percent yield is 33.3%, what is the mass of silver that is actually produced?

16. Considering the equation, 2H2 + O2 → 2H2O
What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2?
Hint: determine limiting reagent first

17. Nitric acid, HNO3, can be manufactured from ammonia, NH3, by using the three reactions
shown below:
Step 1 4 NH3 + 5 O2 → 4 NO + 6 H2O
Step 2 2 NO + O2 → 4 NO2
Step 3 3 NO2 + H2O → NO + 2 HNO3

(Assume that the NO produced in Step 3 is not recycled back into Step 2.) What is the maximum
volume of HNO3 that can be obtained from 4.00 moles of NH3?

9. CHY 30 Mass Relationships in Chemical Reactions 9 of 9