Practice - Stoichiometry
NAME:_____________________________
Show all work and remember to use significant figures and units!
1.
The complete combustion of butane, C4H10, is given by the following equation. Write the products
of the equation and balance it.
_2_ C4H10 (l) + _13_ O2 (g) → _8_ CO2
a.
How many moles of oxygen are needed to burn 10.0 moles of butane in this way?
10.0molC4 H 10
b.
+ _10_ H2O
13molO2
= 65molO2
2molC4 H 10
When 11.3 grams of butane are burned, how many grams of oxygen are needed?
11.3gC4 H 10 1molC4 H 10 13molO2 32.00gO2
= 40.4gO2
58.14gC4 H 10 2molC4 H 10 1molO2
c.
At STP, how many grams of butane would be needed to react with 83.26 mL of oxygen gas?
83.26mLO2
2.
1L
1molO2 2molC4 H 10 58.14gC4 H 10
= 0.03325gC4 H 10
1000mL 22.4LO2 13molO2 1molC4 H 10
Many antacids contain aluminum hydroxide as their active ingredient.
a.
Write the balanced chemical equation for the reaction of hydrochloric acid in stomach acid
with solid aluminum hydroxide to form water and aqueous aluminum chloride.
3 HCl + 1 Al(OH)3 → 3 H2O + 1 AlCl3
b.
How many molecules of hydrochloric acid react with 5.00 grams of aluminum hydroxide?
5.00gAl(OH )3 1molAl(OH )3
3molHCl 6.02x10 23 moleculesHCl
= 1.15x10 23 moleculesHCl
78.01gAl(OH )3 1molAl(OH )3
1molHCl
c.
How many grams of aluminum hydroxide are needed to react with 250. mL of 2.00 M
hydrochloric acid.
250.mL
1L
= 0.250L
1000mL
( 0.250L ) ( 2.00MHCl ) 1molAl(OH )3
3molHCl
78.01gAl(OH )3
= 13.0gAl(OH )3
1molAl(OH )3
3.
Hydrofluoric acid, HF, cannot be stored in glass bottles because compounds called silicates in the
glass are attacked by the HF. For example, sodium silicate, Na2SiO3, reacts in the following way:
_1_ Na2SiO3 (s) + _8_ HF (aq) → _1_ H2SiF6 (aq) + _2_ NaF (aq) + _3_ H2O (l)
a.
How many moles of HF are required to dissolve 2.50 moles of sodium silicate in this reaction?
2.50molNa2 SiO3
b.
8molHF
= 20.0molHF
1molNa2 SiO3
How many grams of sodium fluoride form when 5.00 moles of HF reacts in this way?
5.00molHF 2molNaF 41.99gNaF
= 52.5gNaF
8molHF 1molNaF
c.
4.
How many grams of sodium silicate can be reacted with 12.0 mL of 1.25 M hydrofluoric acid
solution?
12.0mL
= 0.0120L
1000mL
( 0.0120L ) (1.25MHF ) 1molNa2 SiO3 122.07gNa2 SiO3 = 0.230gNa SiO
2
3
8molHF
1molNa2 SiO3
Balancing the following equation!
_1_ Fe3O4 (s) + _4_ CO (g) → _3_ Fe (s) + _4_ CO2 (g)
a.
How many grams of carbon dioxide are produced, if 64.11 grams of Fe3O4 are reacted?
64.11gFe3O4
b.
How many liters of CO are needed to react with 120 grams of Fe3O4 at STP?
120gFe3O4
c.
1molFe3O4
4molCO2 44.01gCO2
= 48.74gCO2
231.55gFe3O4 1molFe3O4 1molCO2
1molFe3O4
4molCO 22.4LCO
= 46LCO
231.55gFe3O4 1molFe3O4 1molCO
How many liters of carbon dioxide are produced if 9.8 liters of CO are used at STP?
9.8LCO2 1molCO2 4molCO 22.4LCO
= 9.8LCO
22.4LCO2 4molCO2 1molCO
d.
How many molecules of CO are needed to react with 16.5 grams of Fe3O4?
16.5gFe3O4
1molFe3O4
4molCO 6.02x10 23 moleculesCO
= 1.72x10 23 moleculesCO
231.55gFe3O4 1molFe3O4
1molCO