Question

A balanced equation for combustion of methane is given below:
$ CH _{4(g)}+2 O _{2(g)} \longrightarrow \\CO _{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 $CH_{4}$ reacts with $2$ moles of oxygen to give one mole of $CO_{2}$ and $2$ moles of water
• B. One molecule of $CH_{4}$ reacts with $2$ molecules of oxygen to give one molecule of $CO_{2}$, and $2$ molecules of water
• C. $22.4 L$ of methane reacts with $44.8 L$ of oxygen to give $44.8 L$ of $CO_{2}$, and $22.4 L$ of water
• D. $16 \,g$ of methane reacts with $64 \,g$ of $O_{2}$, to give $44 \,g$ of $CO_{2}$ and $36\, g$ of water

Answer 1

Answer: (C)

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