Question

if 2.5g of a hydrocarbon that contains 79.95% C and 20.15% H reacts with 2.5g of O2, what is the theoretical yield of CO2 that can form? the molar mass of the hydrocarbon is 30.08g/mol. show work

Answer 1

Answer 2

Calculation:

1. Determine the Molecular Formula:

• Given: 79.95% C, 20.15% H → Empirical formula = CH₃

• Molar mass of CH₃ = 15.03 g/mol

• Given molar mass = 30.08 g/mol → Molecular formula = C₂H₆ (Ethane).

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2. Balanced Reaction:

   $$2{\text{C}}_2{\text{H}}_6+7{\text{O}}_2\to 4{\text{CO}}_2+6{\text{H}}_2\text{O}$$

3. Moles of Reactants:

• Moles of C₂H₆ = $\frac{2.5\text{g}}{30.08\text{g/mol}}=0.0831\text{mol}$

• Moles of O₂ = $\frac{2.5\text{g}}{32.00\text{g/mol}}=0.0781\text{mol}$

4. Limiting Reactant:

• O₂ required for 0.0831 mol C₂H₆ = $0.0831\times \frac{7}{2}=0.291\text{mol}$

• O₂ is limiting (0.0781 mol available).

5. Theoretical Yield of CO₂:

• Moles of CO₂ = $0.0781{\text{mol O}}_2\times \frac{4}{7}=0.0446{\text{mol CO}}_2$

• Mass of CO₂ = $0.0446\text{mol}\times 44.01\text{g/mol}=1.96\text{g}$

Answer:

Theoretical yield of CO₂ = 1.96 grams