2.5 grams of water decomposes into oxygen gas and hydrogen gas. How many grams of hydrogen gas would you expect to produce?

Step-by-step solution

1. 1

   Balance the decomposition first — the mole ratio comes from the coefficients and nothing else. Water splits into diatomic hydrogen and diatomic oxygen:

   $2\,\mathrm{H_2O} \rightarrow 2\,\mathrm{H_2} + \mathrm{O_2}$

   The coefficients $2$ and $2$ give a $1:1$ mole ratio between $\mathrm{H_2O}$ consumed and $\mathrm{H_2}$ produced. That ratio is not $1:1$ by mass, which is why the grams-to-moles detour cannot be skipped.

2. 2

   Collect the molar masses. $M(\mathrm{H_2O}) = 2(1.0) + 16.0 = 18.0\ \mathrm{g/mol}$ and $M(\mathrm{H_2}) = 2(1.0) = 2.0\ \mathrm{g/mol}$.

3. 3

   Convert the 2.5 g of water to moles.

   $n(\mathrm{H_2O}) = \frac{2.5\ \mathrm{g}}{18.0\ \mathrm{g/mol}} = 0.1389\ \mathrm{mol}$

4. 4

   Apply the 1:1 ratio and convert back to mass. One mole of water yields one mole of $\mathrm{H_2}$, so $n(\mathrm{H_2}) = 0.1389\ \mathrm{mol}$ and

   $m(\mathrm{H_2}) = 0.1389 \times 2.0 = 0.278\ \mathrm{g}$

   The whole chain collapses to $2.5 \times \tfrac{2.0}{18.0}$.

5. 5

   Round to two significant figures and check with conservation of mass. The data carry two significant figures, so report $0.28\ \mathrm{g}$. The oxygen produced must then be $2.5 - 0.278 = 2.222\ \mathrm{g}$, and $\tfrac{2.222}{32.0} = 0.0694\ \mathrm{mol}$ of $\mathrm{O_2}$ — exactly half of the $0.1389\ \mathrm{mol}$ of $\mathrm{H_2}$, matching the $2:1$ coefficient ratio in the balanced equation.