Question

Using documented problem solving , calculate the volume of each product and the energy involved in the transformation for the complete combustion of 1,825.0 mL of butane (C4H10) gas and 18.00L of oxygen gas at 760.0 mmHg and 25.0C.

Answer 1

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Combustion reaction of butane is as follows:

$2C_{4}H_{10}_{(g)}+13O_{2}_{(g)}\rightarrow 8CO_{2}_{(g)}+10H_{2}O_{(l)}$

No.of moles of butane,

$n_{butane}=\frac{1.825\;L}{22.4\;L/mol}=0.08147\;mol$

Hence, According to the balanced equation, no.of moles of oxygen required to burn this much butane is given by,

$n_{O_{2}}=\frac{0.08147\;mol}{2}\times 13=0.5296\;mol$

Volume of O₂ used, $V_{O_{2}}=n_{O_{2}}\times22.4\;L/mol=0.5296\;mol\times22.4\;L/mol\mathbf{=11.8625\;L}$

No.of moles of CO₂ produced,

$n_{CO_{2}}=\frac{0.08147\;mol}{2}\times 8=0.32589\;mol$

Volume of CO₂ produced, $V_{CO_{2}}=n_{CO_{2}}\times22.4\;L/mol=0.32589\;mol\times22.4\;L/mol \mathbf{=7.3\;L}$

No.of moles of H₂O produced,

$n_{H_{2}O}=\frac{0.08147\;mol}{2}\times 10=0.4074\;mol$

Volume of H₂O produced, $V_{H_{2}O}=n_{H_{2}O}\times22.4\;L/mol=0.4074\;mol\times22.4\;L/mol \mathbf{=9.125\;L}$

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We know that, Enthalpy change for the reaction is given by,

$\Delta H_{rxn}=\Delta H_{f}(products)-\Delta H_{f}(reactants)$

.

$\Rightarrow \Delta H_{rxn}=\left [\Delta H_{f}(CO_{2}_{(g)})+\Delta H_{f}(H_{2}O_{(g)})\right ]- \left [ \Delta H_{f}(C_{4}H_{10}_{(g)})+\Delta H_{f}(O_{2}_{(g)}) \right ]$

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  $\Rightarrow \Delta H_{rxn}=\left [\(-393.5\;kJ/mol)\;kJ/mol+(-286.3\;kJ/mol)\right ]- \left [ -126\;kJ/mol+(0\;kJ/mol ) \right ]$

$\mathbf{\therefore \Delta H_{rxn}=-553.8\;kJ/mol}$

This is for 1 mol of butane. For n mole of butane heat released is given by:

$\Delta H_{rxn}=-553.8\;kJ/mol\times0.08147\;mol$

$\mathbf{\therefore \Delta H_{rxn}=-45.12\;kJ}$

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