Question

Given,0.2 g of ${H_2}$ reacts with 3.2 g ${O_2}$to form 9 g of water. If 20% of reactant mixture is converted into product. Find a given mass of ${H_2}$, ${O_2}$ and water.

Answer 1

Hint: We have been given the mass of reactant and their percentage. By assuming the total mass to be some variable, we can find out the total mass of reactant each.

Complete answer:
First, let us write what is given to us in the question; then things become easier to do.
Given :
Mass of ${H_2}$= 0.2 g
Mass of ${O_2}$= 3.2 g
Percentage of reactant mixture = 20%
Mass of water formed = 9 g
To find :
given mass of ${H_2}$, ${O_2}$ and water
This question may seem complicated but it is easier and it does not involve any chemical reaction. It just checks our mathematical approach.
We have been given that the reactant that reacted is 20 % of total.
So, if I say that the total mass of ${H_2}$ is ‘x’. Then 20 % of x reacted.
So, we can find as-
20 % of ‘x’ = 0.2 g
$\dfrac{{20}}{{100}}$x = 0.2
‘x’ = $0.2 \times \dfrac{{100}}{{20}}$
‘x’ = 1 g
So, the total mass of ${H_2}$ is 1 g.
Similarly, for ${O_2}$; we can say that the total mass of ${O_2}$ is ‘y’. Then 20 % of y reacted.
Thus, we can find as-
20 % of ‘y’ = 3.2 g
$\dfrac{{20}}{{100}}$y = 3.2
‘y’ = $3.2 \times \dfrac{{100}}{{20}}$
‘y’ = 16 g
So, the total mass of ${O_2}$ is 16 g.
Now, for ${H_2}O$; if the total mass of ${H_2}O$ is ‘z’. Then 20 % of z formed.
So, we can find as-
20 % of ‘z’ = 9 g
$\dfrac{{20}}{{100}}$z = 9
‘z’ = $9 \times \dfrac{{100}}{{20}}$
‘z’ = 45 g
So, the total mass of ${H_2}O$ is 45 g.

Note:
It must be noted that in general, when nothing is mentioned; the one mole of hydrogen gas and one mole of oxygen gas react to give two moles of water. If we discuss in terms of gram masses then 2 g of hydrogen gas reacts with 32 g of oxygen gas to form 36 g of water.