Question

A balanced equation of methane is given below:
$C{H_{4\left( g \right)}} + 2{O_{2\left( g \right)}} \to C{O_{2(g)}} + 2{H_2}{O_{(g)}}$
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. $22.4L$ methane reacts with $44.8L$ of oxygen to give $44.8L$ of $C{O_2}$ and $22.4L$ of water.
D. $16g$ of methane reacts with $64g$ of ${O_2}$ to give $44g$ of $C{O_2}$ and $36g$ of water.

Answer 1

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 $C{H_{4\left( g \right)}} + 2{O_{2\left( g \right)}} \to C{O_{2(g)}} + 2{H_2}{O_{(g)}}$

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 $16g$ of methane reacts with $64g$ of ${O_2}$ to give $44g$ of $C{O_2}$ and $36g$ of water is also correct.
Let's check the last statement which is $22.4L$ methane reacts with $44.8L$ of oxygen to give $44.8L$ of $C{O_2}$ and $22.4L$ of water. We know that the volume of one mole gas at STP is $22.4L$ .So the amount of oxygen required or used for combustion of one mole methane $= 2 \times 22.4L = 44.8L$ and it will give $22.4L$ of carbon dioxide and $44.8L$ 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.