Name:
Answer Key
Honors Chemistry: 2 yellow 2 blue 2 red
Stoichiometry #2 (Limiting Reactant)
1. How many moles of H2 O would be produced if 3.5 mol H2 react with 1.5 mol O2 in the
following reaction:
2 H2 + O2 −→ 2 H2 O
3.5 mol H2 could make 3.5 mol H2 O
1.5 mol O2 could make 3.0 mol H2 O
O2 is limiting, so 3.0. mol H2 O will be produced.
2. If 12.0 moles of S8 reacted with 100. moles of O2 in the unbalanced equation:
S8 +
O2 −→
SO3
Which reactant is limiting, and how much of the other reactant would be left over?
The balanced equation is S8 + 12 O2 −→ 8 SO3
12.0 mol S8 could make 96.0 mol SO3
100. mol O2 could make 66.7 mol SO3
O2 is limiting , and 100. mol O2 will consume 8.33 mol S8
12.0 − 8.33 = 3.7 mol S8 will be left over.
3. 325 g of H2 O is poured onto a 450 g block of sodium metal. What is the limiting reactant?
How many liters of H2 gas are produced at S.T.P.? The equation for this reaction is:
2 Na + 2 H2 O −→ 2 NaOH + H2
1 mol
= 18.0 mol H2 O, which could make 9.00 mol H2
18.0152 g
1 mol
450 gNa ×
= 19.6 mol Na, which could make 9.79 mol H2
22.99 g
325 gH2 O ×
H2 O is limiting , and 9.00 mol H2 was produced.
9.00 mol ×
22.4 `
= 201.6 ` H2
1 mol
4. If 120 g of sand (SiO2 ) is poured into 0.50 ` of 2.0 M hydrofluoric acid (HF), identify the
limiting reactant and determine how much silicon tetrafluoride would be produced? The
chemical equation for this reaction is:
SiO2 + 4 HF −→ SiF4 + 2 H2 O
1 mol
= 2.0 mol SiO2 , which could make 2.0 mol SiF4
60.08 g
2.0 mol
0.50 ` ×
= 1.0 mol HF, which could make 0.25 mol SiF4
`
HF is limiting , and 0.25 mol SiF4 was produced.
120 g SiO2 ×
5. When we blew up the pumpkin on Halloween, we added approximately 5.0 g of H2 O to
approximately 5.0 g of calcium carbide (CaC2 ). Identify the limiting reactant and determine
how many liters of acetylene (C2 H2 ) gas would have been produced at S.T.P., assuming the
unbalanced equation (which you will need to balance first) is:
CaC2 + H2 O −→ C2 H2 + Ca(OH)2
(Remember that H2 O is H+ + OH− in a chemical reaction.)
The balanced equation is CaC2 + 2 H2 O −→ C2 H2 + Ca(OH)2
1 mol
5.0 g H2 O ×
= 0.278 mol H2 O could make 0.139 mol C2 H2
18.0152 g
1 mol
= 0.0780 mol CaC2 could make 0.0780 mol C2 H2
5.0 g CaC2 ×
64.10 g
CaC2 is limiting
0.0780 mol C2 H2 22.4 `
×
= 1.75 ` C2 H2 produced.
1
1 mol
6. 3.27 g of Zn are reacted with 100. m` of 1.00 M HCl in the reaction:
Zn + 2 HCl −→ ZnCl2 + H2
(a) Determine which reactant is limiting.
(b) Determine the number of grams of ZnCl2 that will be produced.
(c) If the reaction conditions are 0◦ C and 1 atm pressure (S.T.P.), determine the number
of liters of H2 gas that will be produced.
(d) If the non-limiting reactant is Zn, determine the mass in grams that will be left over.
(e) If the non-limiting reactant is HCl, determine the concentration (molarity) of the HCl
after the reaction is complete. (Assume the volume does not change.)
5g
1 mol
×
= 0.0765 mol could make 0.0765 mol ZnCl2
1
65.39 g
0.100 ` 1.00 mol
HCl:
×
= 0.100 mol could make 0.0500 mol ZnCl2
1
`
HCl is limiting .
(a) Zn:
g
(b) 0.0500 mol ZnCl2 × 136.30
= 6.82 g ZnCl2
1 mol
22.4 `
(c) 0.0500 mol H2 ×
= 1.12 ` H2
1 mol
(d) The reaction will use 0.0500 mol Zn, which means there will be
0.0765 − 0.0500 = 0.0265 mol Zn left over.
65.39 g
0.0265 mol ×
= 1.73 g Zn left over.
1 mol
(e) N/A
7. Suppose 0.500 m` of hexane (C6 H14 ) (molar mass 86.18 g), which has a density of 0.660 g/m`,
is burned in a combustion reaction inside a sealed Ziploc bag containing 2.00 ` of air at S.T.P.
(The mole fraction of oxygen in air, χO2 = 0.2095. Assume that none of the other gases in
air react with the hexane.)
(a) Write the balanced equation for the combustion reaction that occurs inside the bottle.
(b) How many grams of hexane are in the bottle?
(c) How many moles of hexane are in the bottle?
(d) How many moles of air are in the bottle?
(e) How many moles of oxygen are in the bottle.
(f) Which reacant is limiting?
(g) How much CO2 and how much H2 O will be produced?
(h) After the reaction is complete, the temperature of the bottle is brought back to 0◦ C (to
requotation any additional changes in volume resulting from thermal expansion/contraction
of the gases in the bottle). Is the volume of the bag larger or smaller than it was before
the reaction occurred? (Assume the water produced by the reaction is all liquid or
solid, and has a negligible affect on the volume of the bag.) Explain.
(a) 2 C6 H14 + 19 O2 −→ 12 CO2 + 14 H2 O
0.660 g
(b) 0.500 m` ×
= 0.330 g hexane
1 m`
1 mol
(c) 0.330 g ×
= 0.00383 mol hexane
86.18 g
1 mol
(d) At S.T.P., 2.00 ` ×
= 0.0893 mol air
22.4 `
0.2095 mol O2
(e) 0.0893 mol ×
= 0.0187 mol O2
1 mol air
(f ) 0.00383 mol C6 H14 would produce 0.0230 mol CO2 .
0.0187 mol O2 would produce 0.0118 mol CO2 .
O2 is limiting
(g) 0.0187 mol O2 would produce 0.0118 mol CO2 (as stated above) and 0.0138 mol H2 O
(h) The volume is smaller , because in the reaction, 0.0187 mol O2 is consumed,
and 0.0118 mol CO2 is produced. More moles of O2 were consumed than the
number of moles of CO2 produced.