Apply the law of constant proportions to calculate the mass of oxygen that will be used up for combustion of 5 g of $H_{2}$ gas.

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Solution

The law of constant proportions states that in a chemical substance, the elements are always present in definite proportions by mass.

Combustion of $H_{2}$ gas produces steam.
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$

We can write the following masses of $H_{2}$ , $O_{2}$ and $H_{2} O$ for the reaction:

$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$
2 g 16 g 18g

According to the reaction, 2 g of $H_{2}$ for combustion requires 16 g of $O_{2}$

1 g of $H_{2}$ will require $\frac{16}{2}$ g of $O_{2}$

$∴$ 5 g of $H_{2}$ requires $\frac{16}{2} \times$ 5 g of $O_{2}$ = 40 g

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Apply the law of constant proportions to calculate the mass of oxygen that will be used up for combustion of 5 g of H2 gas.

Apply the law of constant proportions to calculate the mass of oxygen that will be used up for combustion of 5 g of $H_{2}$ gas.