A balanced equation of methane is given below:
$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$
Which of the following statements is not correct on the basis of the above chemical equation?
A. One mole of $C H_{4}$ reacts with $2$ moles of oxygen to give one mole of $C O_{2}$ and $2$ moles of water.
B. One molecule of $C H_{4}$ reacts with $2$ molecules of oxygen to give one molecule of $C O_{2}$ and $2$ molecules of water.
C. $22.4 L$ methane reacts with $44.8 L$ of oxygen to give $44.8 L$ of $C O_{2}$ and $22.4 L$ of water.
D. $16 g$ of methane reacts with $64 g$ of $O_{2}$ to give $44 g$ of $C O_{2}$ and $36 g$ of water.

Question

A balanced equation of methane is given below:
$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$
Which of the following statements is not correct on the basis of the above chemical equation?
A. One mole of $C H_{4}$ reacts with $2$ moles of oxygen to give one mole of $C O_{2}$ and $2$ moles of water.
B. One molecule of $C H_{4}$ reacts with $2$ molecules of oxygen to give one molecule of $C O_{2}$ and $2$ molecules of water.
C. $22.4 L$ methane reacts with $44.8 L$ of oxygen to give $44.8 L$ of $C O_{2}$ and $22.4 L$ of water.
D. $16 g$ of methane reacts with $64 g$ of $O_{2}$ to give $44 g$ of $C O_{2}$ and $36 g$ of water.

Vedantu Content Team · Accepted Answer

Hint : As it is mentioned in the problem that the given equation is balanced so we can easily calculate the number of moles of reactant and products. The given reaction is

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

Complete answer:
> In this reaction we can see that one mole of methane reacts with two moles of oxygen, one mole of carbon dioxide and two moles of water so option A is correct and also option B is correct. Since we know that

1

mole of methane is equal to atomic weight of carbon 4X atomic weight of hydrogen

= 12 + 4 \times 1 = 16

g so we can also calculate mass of oxygen and carbon dioxide. Hence two moles of oxygen

= 2 \times

molecular weight of oxygen

= 2 \times 32 = 64

g.
One mole of carbon dioxide

=

atomic weight of carbon +

2 \times

molecular weight of oxygen

= 12 + 2 \times 32 = 44

g .
Two moles of water

= 2 \times (2 \times

Atomic weight of hydrogen +atomic weight of oxygen)

= 2 \times (2 \times 1 + 16)

= 18

g.
> With the above calculation we can say that the given statement

16 g

of methane reacts with

64 g

of

O_{2}

to give

44 g

of

C O_{2}

and

36 g

of water is also correct.
> Let's check the last statement which is

22.4 L

methane reacts with

44.8 L

of oxygen to give

44.8 L

of

C O_{2}

and

22.4 L

of water. We know that the volume of one mole gas at STP is

22.4 L

.So the amount of oxygen required or used for combustion of one mole methane

= 2 \times 22.4 L = 44.8 L

and it will give

22.4 L

of carbon dioxide and

44.8 L

of water as they have one mole and two mole respectively contribution in the reaction. Hence this statement is wrong.Therefore option D is the correct answer.

Note : We have solved this problem by calculating weights of the reactant and products on the basis of the given reaction. So the answer of this problem is option is C as only this option is not correct on the basis of the given reaction.

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