If 20.0 kg CO₂ are produced from the combustion of butane (C₄H₁₀), calculate the mass of oxygen gas that was consumed.

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The balanced chemical equation for the combustion of butane (C4H10) is:$$2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$$From the equation, 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide. Given that 20.0 kg of $CO_2$ are produced, we first need to find the number of moles of $CO_2$ produced:$$20.0 \, \text{kg} \, CO_2 \times \frac{1000 \, \text{g}}{1 \, \text{kg}} \times \frac{1 \, \text{mole}}{44.01 \, \text{g/mol}} = 454.4 \, \text{moles} \, CO_2$$Since 8 moles of $CO_2$ are produced from 13 moles of $O_2$, we can find the moles of $O_2$ consumed:$$454.4 \, \text{moles} \, CO_2 \times \frac{13 \, \text{moles} \, O_2}{8 \, \text{moles} \, CO_2} = 736.8 \, \text{moles} \, O_2$$Finally, converting moles of $O_2$ to mass:$$736.8 \, \text{moles} \, O_2 \times \frac{32.00 \, \text{g/mol}}{1 \, \text{mole}} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 23.6 \, \text{kg} \, O_2$$Therefore, 23.6 kg of oxygen gas was consumed.

If 200 Kg CO2 Are Produced from the Combustion of Butane

If 200 Kg CO₂ Are Produced from the Combustion of Butane