Chem 2 AP HW 3

Chem 2 AP Homework #3-5: Problems pg. 107 #3.61, 3.66, 3.68, 3.72, 3.74, 3.76, 3.78, 3.106, 3.144
61
Stoichiometry is based on the law of Conservation of Mass. Balanced equations must be used in
order to get the ratios of the participants (reactants and products) in the reaction correct.
66
→ SiCl4(l)
Si(s) + 2Cl2(g) !!
Because the balanced equation is given in the problem, the mole ratio between Cl2 and SiCl4 is
known: 2 moles Cl2 ! 1 mole SiCl4. From this relationship, we have two conversion factors.
2 mol Cl 2
1 mol SiCl 4
and
1 mol SiCl 4
2 mol Cl 2
Use the conversion factor on the left. Moles of SiCl4 will cancel, leaving units of "mol Cl2" for the
answer.
2 mol Cl 2
= 1.01 mol Cl 2
1 mol SiCl 4
Starting with the 5.0 moles of C4H10, we can use the mole ratio from the balanced equation to
calculate the moles of CO2 formed.
? mol Cl 2 reacted = 0.507 mol SiCl 4 ×
68
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
? mol CO2 = 5.0 mol C 4 H 10 ×
72
8 mol CO2
1
= 20 mol CO2 = 2.0 × 10 mol CO2 or 20. mol CO2
2 mol C 4 H 10
→ 2C2H5OH + 2CO2
C6H12O6 !!
glucose
ethanol
The molar mass of glucose = 6(12.01 g) + 12(1.008 g) + 6(16.00 g) = 180.16 g. The molar mass of
ethanol is 2(12.01 g) + 6(1.008g) + 16.00 g = 46.07 g. The balanced equation is given, so the mole
ratio between glucose and ethanol is known; that is 1 mole glucose ! 2 moles ethanol.
grams of glucose → moles of glucose → moles of ethanol → grams of ethanol
? g C 2 H 5OH = 500.4 g C 6 H 12O6 ×
1 mol C6 H 12O6
2 mol C 2 H 5OH
46.07 g C 2 H 5OH
×
×
180.16 g C6 H 12O6
1 mol C6 H 12O6
1 mol C 2 H 5OH
= 255.9 g C2H5OH
Check: Does the answer seem reasonable? Should the mass of ethanol produced be approximately
half the mass of glucose reacted? Twice as many moles of ethanol are produced compared to the
moles of glucose reacted, but the molar mass of ethanol is about one-fourth that of glucose.
The liters of ethanol can be calculated from the density and the mass of ethanol.
volume =
mass
density
Volume of ethanol obtained = 255.9 g × 1 mL
= 324 mL = 0.324 L
0.789 g
2
74
Homework #3-5 Answer key
The balanced equation shows that eight moles of KCN are needed to combine with four moles of
Au.
? mol KCN = 29.0 g Au ×
76
1 mol Au
8 mol KCN
×
= 0.294 mol KCN
197.0 g Au
4 mol Au
→ N2O(g) + 2H2O(g)
(a) NH4NO3(s) !!
(b) Molar mass N2O = 2(14.01 g) + 16.00 g = 44.02 g.
mol NH4NO3 → mol N2O → g N2O
? g N 2O = 0.46 mol NH 4 NO3 ×
78
1 mol N 2O
44.02 g N 2O
×
= 2.0 × 101 g N 2O
1 mol NH 4 NO3
1 mol N 2O
The balanced equation for the decomposition is :
→ 2KCl(s) + 3O2(g)
2KClO3(s) !!
? g O2 = 46.0 g KClO3 ×
106
1 mol KClO3
3 mol O2
32.00 g O2
×
×
= 18.0 g O2
122.6 g KClO3 2 mol KClO3
1 mol O2
The mass of oxygen in MO is 39.46 g − 31.70 g = 7.76 g O. Therefore, for every 31.70 g of M, there
is
7.76 g of O in the compound MO. The molecular formula shows a mole ratio of 1 mole M : 1 mole
O. First, calculate moles of M that react with 7.76 g O.
mol M = 7.76 g O ×
molar mass M =
1 mol O
1 mol M
×
= 0.485 mol M
16.00 g O 1 mol O
31.70 g M
= 65.4 g/mol
0.485 mol M
Thus, the atomic mass of M is 65.4 amu. The metal is most likely Zn.
144
(a) The balanced chemical equation is:
→ 3CO(g) + 7H2(g)
C3H8(g) + 3H2O(g) !!
(b) The following strategy is used to solve this problem. In this problem, we use kg-mol to save a
couple of steps.
kg C3H8 → kg-mol C3H8 → kg-mol H2 → kg H2
? kg H 2 = (2.84 × 103 kg C 3 H 8 ) ×
= 9.09 × 102 kg H2
1 kg-mol C 3 H 8
7 kg-mol H 2
2.016 kg H 2
×
×
44.09 kg C 3 H 8 1 kg-mol C 3 H 8 1 kg-mol H 2

Download Report