# Question #45702

Feb 10, 2016

Solution

#### Explanation:

${H}_{2} + {O}_{2} \setminus \rightarrow {H}_{2} O$

You have 2 atoms of O on the left while you have one on the right side of the equation. Correct that as below

${H}_{2} + {O}_{2} \setminus \rightarrow 2 {H}_{2} O$

Now oxygen molecule is balanced. Take hydrogen. 2 atoms on the left side. We have 4 on the other. Balance this.

$2 {H}_{2} + {O}_{2} \setminus \rightarrow 2 {H}_{2} O$

Now all the elements are balanced in the equation.

As per this equation 2 moles of ${H}_{2}$ and 1 mole of ${O}_{2}$ give 2 moles of water.

1 gmole of ${O}_{2}$ = 32 g of ${O}_{2}$
10 g of ${O}_{2}$ = $\frac{10}{32}$ gmole of ${O}_{2}$
$= 0.3125$ gmole of ${O}_{2}$.

If we have ${H}_{2}$ available at a quantity $\setminus \ge 2 \setminus \times 0.3125$ gmole (i.e) $0.625$ gmole then the product formed will be
${H}_{2} O$ formed$= 2 \setminus \times 0.3125$ gmole
Molecular weight of water $= 18$ g/gmole
${H}_{2} O$ formed$= 2 \setminus \times 0.3125 \cdot 18$ g/gmole *gmole
${H}_{2} O$ formed$= 11.25$ g.

Otherwise the product formed is a function of ${H}_{2}$ availability.