Commercial LPG is a mixture of n-butane and iso-butane. Assuming it to be butane (), what is the app — States of Matter and Gaseous State Chemistry Question
Question
Commercial LPG is a mixture of n-butane and iso-butane. Assuming it to be $100\%$ butane ($C_4H_{10}$), what is the approximate volume of oxygen gas needed to completely burn $1\text{ kg}$ of LPG at NTP?
Answer: B
💡 Solution & Explanation
Reaction: $C_4H_{10} + 6.5O_2 \to 4CO_2 + 5H_2O$. Moles of butane = $1000\text{g} / 58\text{g/mol} = 17.24\text{ mol}$. Moles of $O_2$ needed = $17.24 \times 6.5 = 112.07\text{ mol}$. Volume at NTP = $112.07 \times 22.4\text{ L} \approx 2510\text{ L}$.
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