8 gm of methane is burnt with 4.48L of $O_{2}$ at STP. Find out the volume of $C O_{2}$ gas produced at STP?

2.24 L

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44.8 L

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11.2 L

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33.6 L

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Solution

The correct option is A 2.24 L
Methane combustion reaction can be written as:

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

moles of $O_{2} g a s = \frac{4.48}{22.4} = 0.2 m o l e s$

2 moles of $O_{2}$ produces 1 mole of $C O_{2}$

0.2 moles of $O_{2}$ produces 0.1 mole of $C O_{2}$

Volume of $C O_{2}$ produced $= 0.1 \times 22.4 = 2.24 l i t r e s$

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8 gm of methane is burnt with 4.48L of O2 at STP. Find out the volume of CO2 gas produced at STP?

The correct option is A 2.24 LMethane combustion reaction can be written as:CH4+2O2→CO2+2H2Omoles of O2 gas=4.4822.4=0.2 moles2 moles of O2 produces 1 mole of CO20.2 moles of O2 produces 0.1 mole of CO2Volume of CO2 produced=0.1×22.4=2.24 litres

8 gm of methane is burnt with 4.48L of $O_{2}$ at STP. Find out the volume of $C O_{2}$ gas produced at STP?