Question

A balanced equation for combustion of methane is given below:
$C{H}_4{(}_g)+2O_2{(}_g)\to C{O}_2{(}_g)+2H_{2}O{(}_g)$
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 44.8L of oxygen to give 44.8L 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.

Answer 1

The correct option is C 22.4 L methane reacts with 44.8L of oxygen to give 44.8L of $C{O}_2$ and 22.4 L of water

$C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(g\right)$

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 16 g) of $C{H}_4$ reacts with 2 moles (or 2 molecules or 64 g) of oxygen to give one mole (or 1 molecule or 44 g) of $C{O}_2$ and 2 mole (or 2 molecules or 36 g) 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.