Question

For the chemical equation given below, what is the amount of methane required to produce 72 g of water?
$C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

• A. 32 g
• B. 44 g
• C. 64 g
• D. 16 g

Answer 1

The correct option is A

32 g

Mass of water ($H_{2}O$) to be produced = 72 g.
Gram molecular mass of a compound is the sum of gram atomic mass of all the constituent atoms.
Gram molecular mass of water = 18 g
No. of moles of $H_{2}O$ in 72 g
= $\frac{\text{Mass of water}}{\text{Gram molecular mass of water}}$ = $\frac{72}{18}=4$ mole

For 2 moles of $H_{2}O$ $\to$ 1 mole of methane is required
So, for 4 moles of $H_{2}O$ $\to$ 2 moles of methane is required.
Gram molecular mass of methane = 16 g
The required amount of methane $=2\times 16=32g$.