WKS – Extra
Stoichiometry Review
NAME Answer Key
Period
Date
1) Disilane (Si2H6) is a highly flammable gas (D = 2.70 g/L @ 25°C) that reacts with oxygen gas to form
solid silicon dioxide (SiO2) and water vapor:
2
Si2H6 (g) + 7
O2 (g) → 4
SiO2 (s) + 6
H2O (g)
a. Balance the equation with the lowest whole number coefficients.
b. How many moles of oxygen would be used to completely react with 12.5 moles of disilane?
7 mol O 2
? mol O 2 = 12.5 mol Si2 H 6 ×
= 43.8 mol O 2
2 mol Si2 H 6
c. How many liters of disilane, at 25°C, would be needed to react with sufficient oxygen to produce
50.0 g of silicon dioxide?
? g Si2 H 6 = 50.0 g SiO 2 ×
1 mol SiO 2
×
2 mol Si2 H 6
×
62.22 g Si2 H 6
60.09 g SiO 2 4 mol SiO 2
1 mol Si2 H 6
!###################" !#########" !##########"
0.4160 mol Si2 H 6
0.8321 mol SiO2
×
1L
= 9.59 L Si2 H 6
2.70 g
25.89
d. When 27.1 L disilane @ STP (NOT 25°C) reacts with 70.1 L O2, also at STP, which reactant is
limiting?
1 mol Si2 H 6
1 mol O 2
27.1 L Si2 H 6 ×
= 1.21 mol Si2 H 6 ; 70.1 L O 2 ×
= 3.13 mol O 2
22.4 L Si2 H 6
22.4 L O 2
mol eq. Si2 H 6 =
1.21 mol Si2 H 6
3.13 mol O 2
= 0.605; mol eq. O 2 =
= 0.447
2 mol Si2 H 6
7 mol O 2
0.605 > 0.447, so O 2 is limiting.
e. How many L of the excess reactant remain after the reaction is complete?
2 mol Si2 H 6 22.4 L Si2 H 6
? L Si2 H 6 used = 3.13 mol O 2 ×
×
= 20.0 L Si2 H 6 used
7 mol O 2
1 mol Si2 H 6
!###################"
0.8941 mol Si2 H 6
? L Si2 H 6 remaining = 27.1 L Si2 H 6 present − 20.0 L Si2 H 6 used = 7.1 L Si2 H 6 remaining
f. What is the theoretical yield of solid silicon dioxide from these reactants?
? g SiO 2 = 3.13 mol O 2 ×
4 mol SiO 2
7 mol O 2
!##################"
×
60.09 g SiO 2
1 mol SiO 2
= 107 g SiO 2
1.79 mol SiO2
g. After the reaction, you find that you have collected 97.8 g silicon dioxide. What is the percent yield?
97.8 g SiO 2
% Yield =
× 100% = 91.4%
107 g SiO 2
2) When solid tin(II) oxide reacts with nitrogen trifluoride gas, solid tin(II) fluoride and gaseous dinitrogen
trioxide are produced.
3
SnO (s) + 2
NF3 (g) → 3
SnF2 (s) +
N2O3 (g)
a. Balance the equation with the lowest whole number coefficients.
b. How many moles of tin(II) oxide would be needed to form 29.4 L of N2O3 at STP?
1 mol N 2O3 3 mol SnO
? mol SnO = 29.4 L N 2O3 ×
×
= 3.94 mol SnO
22.4 L N 2O3 1 mol N 2O3
!###################"
1.313 mol N 2O3
c. How many grams of tin(II) oxide would be needed to completely react with 44.1 L of nitrogen
trifluoride (D = 2.90 g/L @ 25°C)?
2.90 g NF3 1 mol NF3
3 mol SnO 134.7 g SnO
? g SnO = 44.1 L NF3 ×
×
×
×
= 364 g SnO
1 L NF3
71.01
g
NF
2
mol
NF
1
mol
SnO
3
3
!###############" !#########" !########"
127.9 g NF3
1.801 mol NF3
2.702 mol SnO
d. When 158.9 g tin(II) oxide reacts with 62.5 g nitrogen trifluoride, which reactant is limiting?
1 mol SnO
1 mol NF3
158.9 g SnO ×
= 1.180 mol SnO; 62.5 g NF3 ×
= 0.880 mol NF3
134.7 g SnO
71.01 g NF3
mol eq. SnO =
1.180 mol SnO
0.880 mol NF3
= 0.3933; mol eq. NF3 =
= 0.440
3 mol SnO
2 mol NF3
0.3933 < 0.440 so SnO is limiting
e. How many grams of the excess reactant remain after the reaction is complete?
2 mol NF3 71.01 g NF3
? g NF3 used = 1.180 mol SnO ×
×
= 55.86 g NF3 used
3 mol SnO 1 mol NF3
!####################"
0.7867 mol NF3
? g NF3 remaining = 62.5 g NF3 present − 55.86 g NF3 used = 6.6 g NF3 remaining
f. What is the theoretical yield of dinitrogen trioxide from these reactants?
1 mol N 2O3 76.02 g N 2O3
? g N 2O3 = 1.180 mol SnO ×
×
= 29.90 g N 2O3
3 mol SnO
1 mol N 2O3
!####################"
0.3933 mol N 2O3
g. After the reaction, you isolate 24.8 g dinitrogen trioxide. What is the percent yield?
24.8 g N 2O3
% Yield =
× 100% = 82.9%
29.90 g N 2O3
Extra Stoichiometry Review WS
2