7- In the reaction,
,
what volume of chlorine, measured at STP, reacts completely with 8 moles of gallium?
A)
B)
C)
D)
E)
Explanation of the answer: In this mole-volume stoichiometry problem # 7 above we are given a balanced
chemical equation and 8 moles of gallium. As we learned in the classroom that the ratios of the coefficients
of a balanced chemical equation are used along with the given moles of one reactant or product to calculate
the number of moles of the unknown and then convert the calculated moles to volume by multiplying it with
22.4 L/mole because one mole of any gas at STP occupies a volume of 22.4 liters.
given 8 moles of gallium X reacts with 3 moles of chlorine
occupies a volume of 22.4 liters
X
2 moles of gallium in the equation
1 mole of chlorine
After cancelling out the cross-multiplying units we are left with
given in choice C. Hence, the choice C is the correct answer
8X 3
2
X 22.4 L; which is exactly what is
8- Acetylene, C2H2, burns completely in O2 to produce CO2 and H2O.
2 C2H2 + 5 O2  4 CO2 + 2 H2O
If 8.30 mole of C2H2 are completely burned in excess oxygen, the number of moles of CO2 produced is
A) 4.00 mol.
B) 4.15 mol.
C) 8.33 mol.
D) 12.5 mol.
E) 16.6 mol.
Explanation of the answer: In this mole-mole stoichiometry problem # 8 above we are given a balanced
chemical equation and 8.30 moles of C2H2. As we learned in the classroom that the ratios of the coefficients
of a balanced chemical equation are used along with the given moles of one reactant or product to calculate
the number of moles of the unknown.
given 8.30 moles of C2H2 X produces 4 moles of carbondioxide
2 moles of C2H2 in the equation
= 16.6 moles of carbondioxide
16.6 mole is exactly what is given in choice E. Hence, the choice E is the correct answer.
9- Mole quantities of product C and reactant A are plotted for a series of experiments. In each case, the
mole quantity of reactant B is 2.00 moles. What are the coefficients in the balanced equation?
?A+?B
?C
A) 0.25, 0.75, 0.50
B) 1.0, 1.0, 1.0
C) 1.0, 1.5, 1.0
D) 1.0, 2.0, 1.0
E) 1.0, 1.0, 2.0
Explanation of the answer: It is obvious from the plotted data that moles of A reactant are plotted on the
X-axis and the moles of the C product are plotted on the Y-axis. Furthermore, the graph also shows that
when 1 mole of reactant A completely reacts with 2 moles of reactant B only I mole of product C is produced
and after that the chemical reaction line becomes parallel to X-axis showing that the amount or number of
moles of product C is not increasing. And that simply means that the chemical reaction is not forming
anymore product either due to the consumption of the limiting reactant or the reaction has reached its
equilibrium point.
In the question, it is given that the reactant B is 2 moles and as we know that the letter B is in the middle of
A and C. Therefore, we should look for the choice which has 2 in the middle. Hence choice D is the correct
answer.
10- In an experiment, 0.0032 mole of maleic acid, C4H4 O4, reacts completely with 0.0064 mole of
sodium hydroxide, NaOH. Which balanced equation describes this reaction?
a.
b.
c.
d.
e.
C4H4O4 + NaOH  NaC4H3O4 + H2O
2 C4H4O4 + NaOH  2 Na2C4H2O4 + 2 H2O
C4H4O4 + 3 NaOH  Na3C4HO4 + 3 H2O
C4H4O4 + 4 NaOH  Na4C4O4 + 4 H2O
C4H4O4 + 2 NaOH  Na2C4H2O4 + 2 H2O
Explanation of the answer: When we divide the given moles of the both reactants with the lower number
0.0032
0.0064
(0.0032), we get 0.0032 = 1 mole of maleic acid, C4H4 O4 and 0.0032 = 2 mole of sodium hydroxide, NaOH.
Equation e given above in the answer choices is the one that has coefficient 1 for maleic acid, C4H4 O4 and 2
for sodium hydroxide, NaOH on the reactant side. Therefore, equation e is the correct answer.
11- What volume of O2 is needed to completely burn 22.4 L of CH4 gas at the same temperature and pressure?
CH4(g) + 2 O2 (g)  CO2 (g) + 2 H2O(g)
A) 11.2 L
B) 2.00 L C) 22.4 L
D) 44.8 L
E) 56.0 L
Explanation of the answer: In this volume-volume stoichiometry problem # 11 above we are given a balanced
chemical equation and 22.4 L of CH4 gas. As we learned in the classroom that the ratios of the coefficients
of a balanced chemical equation are used along with the given volume of one reactant or product to calculate
the volume of the unknown.
given 22.4 L of CH4 gas X cosumes 2 L of O2 gas
1 L of CH4 gas in the equation
= 48.8 L of O2 gas.
Hence, choice D is the correct answer.
12- I. 8 moles of Hydrogen gas contain 4.82 x 1028 atoms
BECAUSE
II. a mole is a measurement unit which contains 6.02 x 1023 particles.
A) I is TRUE, II is FALSE
B) I is FALSE, II is TRUE
C) I and II are BOTH FALSE
D) I and II are BOTH TRUE but II IS NOT a correct explanation of I
E) I and II are BOTH TRUE and II IS a correct explanation of I
Explanation of the answer: In the problem # 12-I above we are given a statement “8 moles of Hydrogen gas
contain 4.82 x 1028 atoms”
Let us see if the statement is correct by calculating the number of atoms in 8 moles of Hydrogen by the
following setup:
given 8 moles of Hydrogen gas X contains 6.02 x 1023 atoms of Hydrogen gas
1 mole of Hydrogen gas
= 48.16 x 1023 = 4.816 x 1024 atoms of Hydrogen
The calculated number of atom is different from the one given in problem 12-I. Therefore, the statement
“8 moles of Hydrogen gas contain 4.82 x 1028 atoms”, made in problem 12-I, is simply false.
However, the statement in 12-II “a mole is a measurement unit which contains 6.02 x 1023 particles.” is true.
Hence, the choice B is the correct answer.
13. Given the reaction:
2 H2 + O2  2 H2O
The total number of grams of O2 needed to produce 54 grams of water is
A) 36
B) 48
C) 61
D) 75
Explanation of the answer: In this mass-mass stoichiometry problem # 13 above we are given a balanced
chemical equation and 54 grams of water and asked to find the total number of grams of O2 needed to produce
54 grams of water.
to solve this problem first we will change the 54 grams of water into moles; then use the calculated moles in
proportion to the moles given in the equation to calculate the mole of O2 and finally convert the calculated
moles of O2 to grams by multiplying it with 32 g, the GFM or MM of O2. as follow:
given 54 g of water
X
equals 1 mole of water
18 g of water
X
used 1 moles of O2
equals 32 g
X
in the equation 2 moles of water GFM or MM of 1 mole of
O2
= 48 g of O2
Hence the correct answer is choice B
14.
Given the reaction:
(NH4)2CO3  2 NH3 + CO2 + H2O
What is the minimum amount of ammonium carbonate that reacts to produce 1.0 mole of ammonia?
A) 0.25 mole
B) 0.50 mole
C) 17 moles
D) 34 moles
Explanation of the answer: In this mole-mole stoichiometry problem # 14 above we are given a balanced
chemical equation and 1 moles of ammonia. As we learned in the classroom that the ratios of the coefficients
of a balanced chemical equation are used along with the given moles of one reactant or product to calculate
the number of moles of the unknown.
given 1 moles of NH3 X 1 mole of ammonium carbonate is needed
in the equation to produce 2 moles of NH3
Hence, the choice B is the correct answer.
1
= 2 mole of ammonium carbonate = 0.50 mole (NH4)2CO3
15.
In a laboratory experiment, a student reacted 2.8 grams of Fe(s) (steel wool) in excess
CuSO4(aq), according to the following balanced equation:
Fe(s) + CuSO4(aq)  FeSO4(aq) + Cu(s)
When the Fe(s) was completely consumed, the precipitated Cu(s) had a mass of 3.2 grams. Did the student's
result in this experiment verify the mole ratio of Fe(s) to Cu(s) as predicted by the equation?
A)
B)
C)
D)
Yes, because the experimental result was 2:1.
No, because the experimental result was 2:1.
Yes, because the experimental result was 1:1.
No, because the experimental result was 1:1.
Explanation of the answer: In this mass-mass stoichiometry problem # 15 above we are given a balanced
chemical equation, 2.8 grams of Fe and 3.2 grams of Cu and asked to find whether it verifies the mole ratio
predicted by the equation.
To solve this problem first we will change the mass given in grams of both metals (Fe and Cu) to moles by
dividing them with atomic mass of each element as follow:
Fe =
Cu =
2.8 𝑔 𝑜𝑓 𝐹𝑒
×
3.2 𝑔 𝑜𝑓 𝐶𝑢
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐹𝑒
56 𝑔 𝑜𝑓 𝐹𝑒
×
= 0.05 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐹𝑒
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑢
63.5 𝑔 𝑜𝑓 𝐶𝑢
= 0. 05 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑢
When we divide both values by 0.05 we get 1 mole of Fe and 1 mole of Cu, which verifies the mole ratio 1:1
predicted in the equation. Hence, the choice C is the correct answer.
Base your answers to questions 16 through 19 on the information below.
A hydrate is a compound that has water molecules within its crystal structure. The formula
for the hydrate CuSO4•5H2O(s) shows that there are five moles of water for every one mole
of CuSO4(s). When CuSO4•5H2O(s) is heated, the water within the crystals is released, as
represented by the balanced equation below.
CuSO4•5H2O(s) —> CuSO4(s) + 5H2O(g)
A student first masses an empty crucible (a heat-resistant container). The student then masses
the crucible containing a sample of CuSO4•5H2O(s). The student repeatedly heats and masses
the crucible and its contents until the mass is constant. The student’s recorded experimental
data and calculations are shown below.
16. Use the student’s data to show a correct numerical setup for calculating the percent
composition by mass of water in the hydrate.
Answer:
17. Identify the total number of significant figures recorded in the calculated mass of CuSO 4•5H2O(s).
Answer: The calculated mass of CuSO 4•5H2O(s) given above 2.13 g has 3 significant figure because the
rule says
if the decimal is present then start from the left and count the first non-zero digit and go all the way to
the right.
18. Explain why the sample in the crucible must be heated until the constant mass is reached.
Answer: Until the samples reaches a constant mass, the student cannot be sure that all the water has been
removed. – to make sure all the water is heated out of the sample.