Questions: If 50.0 g of H2 and 109.0 g of O2 react, how many moles of H2O can be produced in the reaction below? The balanced equation is 2 H2(g) + O2(g) → 2 H2O(g)

Solution Steps

First, we need to calculate the number of moles of $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ using their given masses and molar masses.

The molar mass of $\mathrm{H}_{2}$ is: \[ 2.016 \, \text{g/mol} \]

The molar mass of $\mathrm{O}_{2}$ is: \[ 32.00 \, \text{g/mol} \]

Using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]

For $\mathrm{H}_{2}$: \[ \text{moles of } \mathrm{H}_{2} = \frac{50.0 \, \text{g}}{2.016 \, \text{g/mol}} = 24.80 \, \text{mol} \]

For $\mathrm{O}_{2}$: \[ \text{moles of } \mathrm{O}_{2} = \frac{109.0 \, \text{g}}{32.00 \, \text{g/mol}} = 3.406 \, \text{mol} \]

The balanced chemical equation is: \[ 2 \mathrm{H}_{2}(\mathrm{~g}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \]

From the equation, 2 moles of $\mathrm{H}_{2}$ react with 1 mole of $\mathrm{O}_{2}$. Therefore, we need to compare the mole ratio of the reactants to determine the limiting reactant.

The required moles of $\mathrm{O}_{2}$ for 24.80 moles of $\mathrm{H}_{2}$: \[ \text{required moles of } \mathrm{O}_{2} = \frac{24.80 \, \text{mol} \, \mathrm{H}_{2}}{2} = 12.40 \, \text{mol} \, \mathrm{O}_{2} \]

Since we only have 3.406 moles of $\mathrm{O}_{2}$, $\mathrm{O}_{2}$ is the limiting reactant.

Using the limiting reactant ($\mathrm{O}_{2}$), we can calculate the moles of $\mathrm{H}_{2} \mathrm{O}$ produced. According to the balanced equation, 1 mole of $\mathrm{O}_{2}$ produces 2 moles of $\mathrm{H}_{2} \mathrm{O}$.

\[ \text{moles of } \mathrm{H}_{2} \mathrm{O} = 2 \times \text{moles of } \mathrm{O}_{2} = 2 \times 3.406 \, \text{mol} = 6.812 \, \text{mol} \]

Final Answer