Chemical Equations and
Equation Stoichiometry Review
1. A sample of magnesium carbonate is heated.
a) Write a balanced molecular equation for this reaction (include phases) and indicate
the type of reaction:
MgCO3(s)  MgO(s) + CO2(g)
If exactly 4.75 g of the magnesium carbonate is heated,
b) Calculate the mass of gas produced:
4.75 g MgCO3 x
1molMgCO3
1molCO2
44.0 gCO2
x
x
= 2.50 g CO2
84.3 gMgCO3 1molMgCO3 1molCO2
c) Calculate the volume of gas produced at STP:
2.50 g CO2 x
1molCO2 22.4 LCO2
= 1.27L CO2
x
44.0 gCO2 1molCO2
d) Calculate the number of gas molecules produced:
2.50 g CO2 x
1molCO2 6.02 X 1023 molecules
= 3.42 X 1022 molecules CO2
x
44.0 gCO2
1mol
e) Calculate the mass of the other product:
4.75 g MgCO3 x
1molMgCO3
1molMgO 40.3 gMgO
x
x
= 2.27 g MgO
84.3 gMgCO3 1molMgCO3 1molMgO
f) Calculate the mass of magnesium carbonate that must be heated to produce 75.0L of
the gas at STP:
75.0 L CO2 x
1molCO2 1molMgCO3 84.3gMgCO3
x
x
 283gMgCO3
22.4 LCO2
1molCO2
1molMgCO3
1
2. When nitrogen gas reacts with oxygen gas to produce nitrogen dioxide gas, 33.2 kJ
nitrogen dioxide of heat is absorbed per mole of nitrogen dioxide.
a) Write a BME for this reaction and include phases and the energy term:
N2(g) + 2O2(g) + 66.4 kJ  2NO2(g)
b) What class of reaction is this:
synthesis
c) Identify substance oxidized and substance reduced:
N2 is oxidized (0  +4); O2 is reduced (0  -2)
If one was to produce 200.0 g of nitrogen dioxide gas,
d) Calculate the masses of nitrogen and oxygen gases needed to react produce this
amount of gas:
200.0 gNO2 x
1molNO2
1molN 2 28.0 gN 2
x
x
 60.87 gN 2
46.0 gNO2 2molNO2 1molN 2
200.0 gNO2 x
1mo l NO2 2molO2 32.0 gO2
x
x
 139.1gO2
46.0 gNO2 2molNO2 1molO2
e) Calculate the energy involved in the reaction in (b):
200.0 gN 2 x
1molNO2
66.4kJ
x
 144.3kJ
46.0 gNO2 2molNO2
f) Calculate the volumes of nitrogen and oxygen gases, measured at STP, needed to
produce 200.0 g of nitrogen dioxide:
200.0 gNO2 x
1molNO2
1molN 2 22.4 LN 2
x
x
 48.70 LN 2
46.0 gNO2 2molNO2 1molN 2
200.0 gNO2 x
1molNO2
2molO2 22.4 LO2
x
x
 97.40 LO2
46.0 gNO2 2molNO2 1molO2
g) Calculate the volume the nitrogen dioxide would occupy at STP:
200.0 gNO2 x
1molNO2 22.4 LNO2
x
 97.4 LNO2
46.0 gNO2 1molNO2
2
3. A sample of potassium metal is placed in water.
a) What class of reaction is this? Single replacement
b) Include phases for:
BME: 2K(s) + 2HOH(l)  2KOH(aq) + H2(g)
CIE:
2K(s) + 2HOH(l)  2K+(aq) + 2OH-(aq) + H2(g)
NIE:
same as CIE
c) Identify substance oxidized and substance reduced:
K is oxidized (0  +1); H+ in HOH is reduced (+1  0 in H2)
If exactly 0.220 g of potassium reacts with excess water,
d) Calculate the mass of gas produced:
0.220 gKx
1molK 1molH 2 2.0 gH 2
x
x
 0.00563gH 2
39.1gK 2molK 1molH 2
e) Will the resulting solution conduct electricity? Explain.
Yes because KOH(aq) consists of freely moving K+(aq) ions and OH-(aq) ions
f) If the solution is allowed to evaporate, calculate the mass of solid left:
0.220 gKx
1molK 2molKOH 56.1gKOH
x
x
 0.316 gKOH (s)
39.1gK
2molK
1molKOH
g) Calculate the mass of water needed to react with the 0.200 g of potassium:
0.200 gKx
1molK 2molHOH 18.0 gHOH
x
x
 0.921gHOH
39.1gK
2molK
1molHOH
h) If the reaction releases 3750 kJ per mole of potassium reacted, calculate the amount
of heat released when the 0.200 g of potassium reacts with the water:
0.200 gKx
1molK 3750kJ
x
 19.2kJ
39.1gK 1molK
3
4. Aqueous solutions of iron(II)chloride and sodium hydroxide are mixed:
a) Include phases for:
BME:
CIE:
NIE:
FeCl2(aq) + 2NaOH(aq)  Fe(OH)2(s) + 2NaCl(aq)
Fe2+(aq) + 2Cl-(aq) + 2Na+(aq) + 2OH-(aq)  Fe(OH)2(s) + 2Na+(aq) + 2Cl-(aq)
Fe2+(aq) + 2OH-(aq)  Fe(OH)2(s)
b) What class of reaction is this?
double replacement
If 3.20 g of the iron(II) chloride reacts with 1.75 g of the sodium hydroxide:
c) Determine the limiting reactant:
3.20 gFeCl2 x
1molFeCl2 1molFe(OH ) 2 89.9 gFe(OH ) 2
x
x
 2.27 gFe(OH ) 2
126.9 gFeCl2
1molFeCl2
1molFe(OH ) 2
1.75 gNaOHx
1molNaOH 1molFe(OH ) 2 89.9 gFe(OH ) 2
x
x
 1.97 gFe(OH ) 2
40.0 gNaOH 2molNaOH
1molFe(OH ) 2
Therefore NaOH is limiting reactant
d) Calculate the mass of precipitate formed: 1.97 g Fe(OH)2
e) Calculate the mass of excess reactant left over:
1.75 gNaOHx
1molNaOH 1molFeCl2 126.9 gFeCl2
x
x
 2.76 gFeCl2 reacts
40.0 gNaOH 2molNaOH 1molFe(OH ) 2
Therefore 3.20g – 2.76 g = 0.44 g FeCl2 left over unreacted
f) If 2.50 g of precipitate is actually collected, calculate the percent yield:
2.50
x100%  127%
1.97
4
5. A sample of gaseous butane, C4H10, is burned in air. The reaction releases 2, 430 kJ per
mole of butane.
a) What class or reaction is this? combustion
b) Write a BME and include the phases and energy term:
2C4H10(l) + 13O2(g)  8CO2(g) + 10H2O(g) + 4860kJ
If exactly 500.g of butane reacts with 1.200 X 103 liters of oxygen gas at STP,
c) Determine the limiting reactant:
500.gC4 H10 x
1molC4 H10
8molCO2 44.0 gCO2
x
x
 1520 gCO2
58.0 gC4 H10 2molC4 H10 1molCO2
1.200 X 103 LO2 x
1molO2 8molCO2 44.0 gCO2
x
x
 1450 gCO2 LR
22.4 LO2 13molO2 1molCO2
d) Calculate the mass of carbon dioxide produced: 1450g CO2
e) Calculate the mass of water produced:
1.200 X 103 LO2 x
1molO2 10molH 2O 18.0 gH 2O
x
x
 742 gH 2O
22.4 LO2 13molO2
1molH 2O
f) Calculate the mass of excess reactant that remains:
1.200 X 103 LO2 x
1molO2 2molC4 H10 58.0 gC4 H10
x
x
 478 gC4 H10 reacts
22.4 LO2 13molO2
1molC4 H10
Therefore start 500.g – 478 g reacts = 22 g C4H10 left
g) Calculate the number of molecules of excess reactant:
1molC4 H10 6.02 X 1023 moleculesC4 H10
22 gC4 H10 x
x
 2.28 X 1023 moleculesC4 H10
58.0 gC4 H10
1molC4 H10
h) Calculate the amount of heat released:
1.200 X 103 LO2 x
1molO2
4680kJ
x
 19300kJ
22.4 LO2 13molO2
5
6. Equal volumes of equal concentrations of sulfuric acid and sodium hydroxide are mixed.
The reaction has ∆H = -53.3 kJ/mol acid. Write the following and include phases:
a) BME: H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2HOH(l) + 53.3kJ
b) CIE: 2H+(aq) + SO42-(aq) +2Na+(aq) + 2OH-(aq)  2Na+(aq) + SO42-(aq) + 2HOH(l)
c) NIE: H+(aq) + OH-(aq)  HOH(l)
d) Classify reaction: double replacement (acid-base)
If exactly 49.0 g of sulfuric acid reacts with sufficient sodium hydroxide,
e) Calculate the mass of water formed:
49.0 gH 2 SO4 x
1molH 2 SO4 2molHOH 18.0 gHOH
x
x
 18.0 gHOH
98.1gH 2 SO4 1molH 2 SO4 1molHOH
f) Name the salt formed: sodium sulfate
g) Calculate the mass of the salt formed:
49.0 gH 2 SO4 x
1molH 2 SO4 1molNa2 SO4 142.1gNa2 SO4
x
x
 71.0 gNa2 SO4
98.1gH 2 SO4 1molH 2 SO4
1molNa2 SO4
h) Calculate the mass of sodium hydroxide reacted:
49.0 gH 2 SO4 x
1molH 2 SO4 2molNaOH 40.0 gNaOH
x
x
 40.0 gNaOH
98.1gH 2 SO4 1molH 2 SO4 1molNaOH
i) Calculate the amount of heat involved in the reaction:
49.0 gH 2 SO4 x
1molH 2 SO4
53.3kJ
x
 26.6kJ
98.1gH 2 SO4 1molH 2 SO4
6
7. A sample of lithium chlorate is heated.
a) Write a BME and include phases:
2LiClO3(s)  2LiCl(s) + 3O2(g)
b) Classify reaction: decomposition
If 1.20 g of the lithium chlorate is heated,
c) Calculate the mass of salt formed:
1.20 gLiClO3x
1molLiClO3
2molLiCl
42.4 gLiCl
x
x
 0.563gLiCl
90.4 gLiClO3 2molLiClO3 1molLiCl
d) Calculate the mass of gas produced:
1.20 gLiClO3 x
1molLiClO3
3molO2
32.0 gO2
x
x
 0.637 gO2
90.4 gLiClO3 2molLiClO3 1molO2
or 1.20 g = 0.563g + xg, x = 0.637g
e) Calculate the volume of gas produced at STP:
0.637 gO2 x
1molO2 22.4 LO2
x
 0.446 LO2
32.0 gO2 1molO2
f) Calculate the number of gas molecules produced:
0.637 gO2 x
1molO2 6.02 X 1023 moleculesO2
x
 1.20 X 1022 moleculesO2
32.0 gO2
1molO2
g) The salt formed does not conduct electricity in the solid phase but will conduct electricity
when dissolved in water. Explain:
In the phase the Li+ and Cl- ions are held in a fixed lattice position and are not free to
move and will not conduct electricity; when dissolved in water, the ions dissociate and
become freely moving hydrated ions and will conduct electricity
7
8. A sample of molten (liquid) potassium chloride undergoes electrolysis.
a) Write a BME and include phases:
2KCl(l)  2K(s) + Cl2(g)
b) Classify reaction: decomposition
c) Calculate the mass of salt that must be electrolyzed to produce 95.0 L of the product
gas at STP:
95.0 LCl2 x
1molCl2 2molKCl 74.6 gKCl
x
x
 633gKCl
22.4 LCl2 1molCl2 1molKCl
d) If 55.0 g of the potassium chloride is electrolyzed, calculate the mass of solid product
formed:
55.0 gKClx
1molKCl
2molK 39.1gK
x
x
 28.8 gK
74.6 gKCl 2molKCl 1molK
e) If 24.6 g of the solid is actually obtained in part (d), calculate the percent yield:
24.6
x100%  85.4%
28.8
f) In a separate electrolysis, 90.0 L of the gas was collected at STP. If this represents a
77.5% yield, calculate the mass of potassium chloride that had to be electrolyzed to
produce 100% yield.
90.0 L
XL

; X  116 L
77.5% 100%
116 LCl2 x
1molCl2 2molKCl 74.6 gKCl
x
x
 773gKCl
22.4 LCl2 1molCl2 1molKCl
8