Questions: Hydrogen (H₂) gas and oxygen (O₂) gas react to form water (H₂O) vapor. Suppose you have 1.0 mol of H₂ and 9.0 mol of O₂ in a reactor. Calculate the largest amount of H₂O that could be produced. Round your answer to the nearest 0.1 mol.

Solution Steps

The balanced chemical equation for the reaction between hydrogen gas and oxygen gas to form water vapor is: \[ 2 \mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} \]

To find the limiting reactant, we need to compare the mole ratio of the reactants with the coefficients in the balanced equation.

From the balanced equation, 2 moles of \(\mathrm{H}_2\) react with 1 mole of \(\mathrm{O}_2\).

Given:

• 1.0 mol of \(\mathrm{H}_2\)
• 9.0 mol of \(\mathrm{O}_2\)

Calculate the required moles of \(\mathrm{O}_2\) for 1.0 mol of \(\mathrm{H}_2\): \[ \text{Required } \mathrm{O}_2 = \frac{1.0 \text{ mol } \mathrm{H}_2}{2} = 0.5 \text{ mol } \mathrm{O}_2 \]

Since we have 9.0 mol of \(\mathrm{O}_2\), which is more than enough, \(\mathrm{H}_2\) is the limiting reactant.

Using the limiting reactant (\(\mathrm{H}_2\)), we can calculate the amount of \(\mathrm{H}_2\mathrm{O}\) produced.

From the balanced equation, 2 moles of \(\mathrm{H}_2\) produce 2 moles of \(\mathrm{H}_2\mathrm{O}\).

Therefore, 1.0 mol of \(\mathrm{H}_2\) will produce: \[ \text{Produced } \mathrm{H}_2\mathrm{O} = 1.0 \text{ mol } \mathrm{H}_2 \times \frac{2 \text{ mol } \mathrm{H}_2\mathrm{O}}{2 \text{ mol } \mathrm{H}_2} = 1.0 \text{ mol } \mathrm{H}_2\mathrm{O} \]

Final Answer