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Q.

Given the oxidation of methane in the presence of oxygen $C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$ . When 8 g of methane is completely oxidized in the presence of oxygen, the volume of $C O_{2}$ formed at STP will be

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a

16.8 liters

b

5.6 liters

c

11.2 liters

d

22.4 liters

answer is B.

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Detailed Solution

C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}

1 mole of $C H_{4}$ - 1 mole of $C O_{2}$

16g of $C H_{4}$ - $22.4 l t o f C O_{2}$

8g of $C H_{4}$ - $11.2 l t o f C O_{2}$

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