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The amount of methane gas is required to obtain 5.6 L of ${CO}_{2}$ gas at STP from the combustion reaction with excess amount of Oxygen gas is present ?

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8 g

32 g

4 g

12 g

answer is C.

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Detailed Solution

${CH}_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

${1 mole of CH}_{4} {---- 1 mole of CO}_{2}$

$16 {g of CH}_{4} {----- 22.4 lit of CO}_{2}$

$? {X ---- 5.6 lit of CO}_{2}$

$16 \times 5.6 = 22.4 \times X$

$X = \frac{16 \times 5.6}{22.4}$

$= 4 {g of CH}_{4}$

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