Question

$4gm$ of hydrogen are ignited with $4gm$ of oxygen. The weight of water formed is:
(A) $0.5gm$
(B) $3.5gm$
(C) $4.5gm$
(D) $2.5gm$

Answer 1

Hint: For solving this problem, we will require a stoichiometric equation of formation of water that is,
${H_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right)$

Complete step by step answer:
Law of definite proportions states that a given chemical compound always contains its components (elements) in fixed ratio (by mass) such as water, it always combines in the ratio of $1:8$.
The given problem is an example of the law of definite proportions.
The balanced equation of formation of water is:
$2{H_2} + {O_2} \to 2{H_2}O$
It means that $2$ moles of hydrogen reacts with one mole of oxygen to produce $2$ moles of water as products.
Similarly, $4g$ of hydrogen gas which means $4g$ of hydrogen $ = \dfrac{4}{2} = 2$ moles will react with $32g$ of oxygen to produce $36g$ of water.
In the question, there is only $4g$ of oxygen that reacts with $4g$of hydrogen. Hydrogen is a limiting reagent.
$4g$ of oxygen $ = \dfrac{4}{{32}} = \dfrac{1}{8}$ moles
$4g$ of hydrogen$ = \dfrac{4}{2} = 2$ moles
So, the water produced $ = \dfrac{4}{{32}} \times 2 = \dfrac{1}{4}$ moles of ${H_2}O$
Weight of ${H_2}O$ formed is $ = \dfrac{4}{{32}} \times 36 = 4.5g$

So, the correct answer is Option C.

Note:
The limiting reagent is the one that is totally consumed during the reaction. So, here in this case ${O_2}$will be the limiting reagent. The product formed in the reaction is limited because of limiting reagent.