Answer:
1) <u>16.8 L CO2</u>
<u>2) 36.96 L NH3</u>
<u>3) </u> <u>9.88 L CO2 </u>
<u>4) 56.99 L H2O</u>
Explanation:
How many liters of carbon dioxide gas will be produced when 75.0 g of calcium carbonate decomposes to form calcium oxide when at STP?
CaCO3 → CaO + CO2
Moles calcium carbonate = 75.0 grams / 100.09 g/mol
Moles calcium carbonate = 0.750 moles
For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2
For 0.750 moles CaCO3 we'll have 0.750 moles CO2
1 mol = 22.4 L
0.750 moles CO2 = 0.750 *22.4 L =<u> 16.8 L CO2</u>
2. Hydrogen gas reacts with 23.1 g of nitrogen gas to produce ammonia (NH3). What volume of ammonia will be produced at STP?
3H2 + N2 → 2NH3
Moles N2 = 23.1 grams / 28.0 g/mol
Moles N2 = 0.825 moles
For 3 moles H2 we need 1 mol N2 to produce 2 moles NH3
For 0.825 moles N2 we'll have 2*0.825 = 1.65 moles NH3
1 mol = 22.4 L
1.65 mol = 1.65 * 22.4 L = <u>36.96 L NH3</u>
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3. Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide. How many liters of carbon dioxide will be produced from 23.5 g of iron (III) oxide when at STP?
Fe2O3 + 3CO → 2Fe + 3CO2
Moles Fe2O3 = 23.5 grams / 159.69 g/mol
Moles Fe2O3 = 0.147 moles
For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2
For 0.147 moles Fe2O3 we'll have 3*0.147 = 0.441 moles CO2
1 mol = 22.4 L
0.441 moles = 22.4 * 0.441 = <u>9.88 L CO2 </u>
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4.How many liters of water vapor would be produced in the combustion of 12.5L of ethane, C2H6 at STP?
2C2H6 + 7O2 →4CO2 + 6H2O
22.4 L = 1 mol
12.5 L = 0.848 moles C2H6
For 2 moles C2H6 we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
For 0.848 moles C2H6 we'll have 3*0.848 = 2.544 moles H2O
1 mol = 22.4 L
2.544 moles = 22.4 L * 2.544 = <u>56.99 L H2O</u>
