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.

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.
Transcript text: Hydrogen $\left(\mathrm{H}_{2}\right)$ gas and oxygen $\left(\mathrm{O}_{2}\right)$ gas react to form water $\left(\mathrm{H}_{2} \mathrm{O}\right)$ vapor. Suppose you have 1.0 mol of $\mathrm{H}_{2}$ and 9.0 mol of $\mathrm{O}_{2}$ in a reactor. Calculate the largest amount of $\mathrm{H}_{2} \mathrm{O}$ that could be produced. Round your answer to the nearest 0.1 mol.
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Solution

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Solution Steps

Step 1: Write the Balanced Chemical Equation

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

Step 2: Determine the Limiting Reactant

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.

Step 3: Calculate the Amount of \(\mathrm{H}_2\mathrm{O}\) Produced

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

\[ \boxed{1.0 \text{ mol}} \]

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