Write and balance the chemical equation for the combustion of butane.

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  1. Write and balance the chemical equation for the combustion of butane.

    Butane (C4H10\text{C}_4\text{H}_{10}) burns in oxygen (O2\text{O}_2) to produce carbon dioxide (CO2\text{CO}_2) and water (H2O\text{H}_2\text{O}).

    2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)2\text{C}_4\text{H}_{10}(g) + 13\text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 10\text{H}_2\text{O}(g)

  2. Calculate the molar masses of the reactants and products involved.

    • Molar mass of C4H10=(4×12.01)+(10×1.008)=48.04+10.08=58.12 g/mol\text{C}_4\text{H}_{10} = (4 \times 12.01) + (10 \times 1.008) = 48.04 + 10.08 = 58.12 \text{ g/mol}
    • Molar mass of O2=2×16.00=32.00 g/mol\text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}
    • Molar mass of CO2=12.01+(2×16.00)=12.01+32.00=44.01 g/mol\text{CO}_2 = 12.01 + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \text{ g/mol}

  3. Convert the given masses of reactants to moles.

    • Moles of C4H10=4.65 g58.12 g/mol=0.07999 mol\text{C}_4\text{H}_{10} = \frac{4.65 \text{ g}}{58.12 \text{ g/mol}} = 0.07999 \text{ mol}
    • Moles of O2=10.0 g32.00 g/mol=0.3125 mol\text{O}_2 = \frac{10.0 \text{ g}}{32.00 \text{ g/mol}} = 0.3125 \text{ mol}

  4. Determine the limiting reactant.

    Using the stoichiometric ratio from the balanced equation (2 mol C4H102 \text{ mol C}_4\text{H}_{10} reacts with 13 mol O213 \text{ mol O}_2):

    • If C4H10\text{C}_4\text{H}_{10} is limiting, moles of O2\text{O}_2 needed = 0.07999 mol C4H10×13 mol O22 mol C4H10=0.5199 mol O20.07999 \text{ mol C}_4\text{H}_{10} \times \frac{13 \text{ mol O}_2}{2 \text{ mol C}_4\text{H}_{10}} = 0.5199 \text{ mol O}_2. Since we only have 0.3125 mol O20.3125 \text{ mol O}_2, oxygen is the limiting reactant.

  5. Calculate the mass of carbon dioxide produced using the limiting reactant (O2\text{O}_2).

    From the balanced equation, 13 mol O213 \text{ mol O}_2 produces 8 mol CO28 \text{ mol CO}_2.

    • Moles of CO2\text{CO}_2 produced = 0.3125 mol O2×8 mol CO213 mol O2=0.1923 mol CO20.3125 \text{ mol O}_2 \times \frac{8 \text{ mol CO}_2}{13 \text{ mol O}_2} = 0.1923 \text{ mol CO}_2
    • Mass of CO2\text{CO}_2 produced = 0.1923 mol×44.01 g/mol=8.463 g0.1923 \text{ mol} \times 44.01 \text{ g/mol} = 8.463 \text{ g}

  6. Calculate the mass of the excess reactant (C4H10\text{C}_4\text{H}_{10}) remaining.

    First, find the moles of C4H10\text{C}_4\text{H}_{10} that reacted with the limiting reactant.

    • Moles of C4H10\text{C}_4\text{H}_{10} reacted = 0.3125 mol O2×2 mol C4H1013 mol O2=0.04808 mol C4H100.3125 \text{ mol O}_2 \times \frac{2 \text{ mol C}_4\text{H}_{10}}{13 \text{ mol O}_2} = 0.04808 \text{ mol C}_4\text{H}_{10}
    • Mass of C4H10\text{C}_4\text{H}_{10} reacted = 0.04808 mol×58.12 g/mol=2.795 g0.04808 \text{ mol} \times 58.12 \text{ g/mol} = 2.795 \text{ g}
    • Mass of C4H10\text{C}_4\text{H}_{10} remaining = Initial mass - Mass reacted = 4.65 g−2.795 g=1.855 g4.65 \text{ g} - 2.795 \text{ g} = 1.855 \text{ g}

    Rounding to three significant figures:

    • Mass of carbon dioxide produced: 8.46 g\boxed{\text{8.46 g}}
    • Mass of excess reactant (C4H10\text{C}_4\text{H}_{10}) remaining: 1.86 g\boxed{\text{1.86 g}}
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