Question

Question #45702

Answer 1

Solution

Explanation:

`H_2 + O_2 \rightarrow H_2O`

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 \rightarrow 2H_2O`

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

`2H_2 + O_2 \rightarrow 2H_2O`

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` = `10/32` gmole of `O_2`
`= 0.3125` gmole of `O_2`.

If we have `H_2` available at a quantity `\ge 2 \times 0.3125` gmole (i.e) `0.625` gmole then the product formed will be
`H_2O` formed`= 2 \times 0.3125` gmole
Molecular weight of water `=18` g/gmole
`H_2O` formed`= 2 \times 0.3125 * 18` g/gmole *gmole
`H_2O` formed`= 11.25` g.

Otherwise the product formed is a function of `H_2` availability.