Question

What is the product of electrolysis of $45$ grams of water?

• A. $40$ grams of $H_2$ and $45$ grams of $O_2$
• B. $40$ grams of $O_2$ and $5$ grams of $H_2$
• C. $30$ grams of $H_2$ and $15$ grams $O_2$
• D. $30$ grams of $O_2$ and $15$ grams of $H_2$

Answer 1

Answer: (B)

$40$ grams of $O_2$ and $5$ grams of $H_2$

The reaction depicting electrolysis of water is as follows:

$2H_{2}O\to 2H_2+O_2$

Molecular weight of water = 18, hydrogen = 2, and oxygen = 32.
$36$ g of $H_{2}O$ produce ---- $4$ g of $H_2$

Applying unitary method,
$45$ g of $H_{2}O$ produce ---- $\frac{4}{36}\times 45$ = $5$ g of $H_2$.
$36$ g of $H_{2}O$ produce ---- $32$ gms of $O_2$.
$45$ g of $H_{2}O$ produce ---- $\frac{32}{36}\times 45$ = $40$ gms of $O_2$.