Question

A balanced equation for combustion of methane is given below:
$$\begin{gathered} C{H}_{4(g)}+2O_{2(g)}⟶ \\ C{O}_{2(g)}+2H_2{O}_{(g)} \end{gathered}$$
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. $22.4L$ of 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

Answer: (C)

$C{H}_4(g)+2O_2(g)\to C{O}_2(g)+2H_{2}O(g)$
Since 1 mole methane $=12+4\times 1=16g$ and $2$ mole oxygen
$=2\times 32=64g,1$ mole
$C{O}_2=12+32=44g$ and 2 mole water $=2\times 18=36g$
From the balanced reaction: One mole(or 1 molecule or $16g$ ) of $C{H}_4$ reacts with 2 moles (or 2 molecules or $64g$ ) of oxygen to give one mole (or 1 molecule or $44g$ ) of $C{O}_2$ and 2 mole (or 2 molecules or $36g$ ) of water.
For the gaseous system at STP:
$22.4L$ methane reacts with $44.8L$ of oxygen to give $22.4L$ of $C{O}_2$ and $44.8L$ of water.