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Mole Concept

CH4+2O2 → CO2...

Question

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
Initially, $16 g$ of $C H_{4}$ is combusted in $22.4 L$ of $O_{2}$ at STP. Identify the correct statement.

C H_{4}

is the limiting reagent

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O_{2}

is the limiting reagent

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C H_{4}

will be completely consumed

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1 mole of

C H_{4}

remain unreacted

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Solution

The correct option is B $O_{2}$ is the limiting reagent
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

1 mole of $C H_{4}$ requires 2 moles of $O_{2}$ for complete combustion.
Molar mass of methane = $16 g$ .
Number of moles of methane in $16 g$ sample = $\frac{16 g}{16 g}$ = $1 mol$ .

$1$ mole of all the gases occupy $22.4 L$ at STP.
Number of moles of oxygen $= 22.4 L = 1 mol (given)$

As only 1 mol of $O_{2}$ is available, $O_{2}$ acts as the limiting reagent in this reaction.

Suggest Corrections

Similar questions

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

16 g

C H_{4}

22.4 L

O_{2}

C O_{2}

16

C H_{4}

48

O_{2}

C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O

The heat of combustion of methane at 298 K is expressed by:

CH4(g)+2O2(g)→CO2(g)+2H2O(l);

ΔH=-890.2 KJ ΔE of the reaction at this temperature is?

11.2 L

C H_{4}

22.4 L

C_{2} H_{6}

11.2 L

C H_{4}

22.4 L

C_{2} H_{6}

S T P

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