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Exercise · Q24

Q.Methane burns in oxygen as CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l). Calculate the volume of CO2\text{CO}_2 gas produced at STP by the complete combustion of 8.0 g8.0\ \text{g} of methane (molar mass CH4=16.04 g mol−1\text{CH}_4 = 16.04\ \text{g mol}^{-1}).

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Balanced equation: CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) — the mole ratio of CH4\text{CH}_4 to CO2\text{CO}_2 is 1:11:1.

Moles of CH4=8.016.04=0.4988 mol\text{CH}_4 = \dfrac{8.0}{16.04} = 0.4988\ \text{mol}

Since the ratio is 1:11:1:

n(CO2)=n(CH4)=0.4988 moln(\text{CO}_2) = n(\text{CH}_4) = 0.4988\ \text{mol}

Volume of CO2\text{CO}_2 at STP: …

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