Question

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Answer 1

Answer:

1) 16.8 L CO2

2) 36.96 L NH3

3) 9.88 L CO2

4) 56.99 L H2O

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 = 16.8 L CO2

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 = 36.96 L NH3

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 = 9.88 L CO2

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 = 56.99 L H2O