Atoms and Molecules - Law of Constant Proportions

$3.0 \text{ g}$ of hydrogen gas reacts with $24.0 \text{ g}$ of oxygen gas to form water. What mass of oxygen gas would be required to react completely with $9.0 \text{ g}$ of hydrogen gas?

According to the Law of Constant Proportions, hydrogen and oxygen always react in a fixed ratio of $1:8$ by mass to form water.

Given: $\text{Mass of Hydrogen} = 3.0 \text{ g}$ $\text{Mass of Oxygen} = 24.0 \text{ g}$ $\text{Ratio} = \frac{3.0}{24.0} = \frac{1}{8}$

To find the oxygen required for $9.0 \text{ g}$ of hydrogen: $\frac{\text{Mass of Hydrogen}}{\text{Mass of Oxygen}} = \frac{1}{8}$ $\frac{9.0}{\text{Mass of Oxygen}} = \frac{1}{8}$ $\text{Mass of Oxygen} = 9.0 \times 8 = 72.0 \text{ g}$

Since the ratio of $H:O$ is constant at $1:8$, if we triple the amount of hydrogen from $3\text{ g}$ to $9\text{ g}$, the required oxygen must also be tripled ($24 \times 3$) or calculated via the fixed ratio.

In an experiment, $1.24 \text{ g}$ of Phosphorus reacted with Oxygen to form $2.84 \text{ g}$ of Phosphorus oxide. In another experiment, $2.48 \text{ g}$ of Phosphorus gave $5.68 \text{ g}$ of the same oxide. Show that these results follow the Law of Constant Proportions.

In Experiment 1: $\text{Mass of P} = 1.24 \text{ g}$ $\text{Mass of Oxide} = 2.84 \text{ g}$ $\text{Mass of O} = 2.84 - 1.24 = 1.60 \text{ g}$ $\text{Ratio of P:O} = \frac{1.24}{1.60} = 0.775$

In Experiment 2: $\text{Mass of P} = 2.48 \text{ g}$ $\text{Mass of Oxide} = 5.68 \text{ g}$ $\text{Mass of O} = 5.68 - 2.48 = 3.20 \text{ g}$ $\text{Ratio of P:O} = \frac{2.48}{3.20} = 0.775$

Since the ratio is the same ($0.775:1$), the law is verified.

The ratio of the masses of the constituent elements (Phosphorus and Oxygen) remains constant across different experiments for the same compound.