Question

Which is the limiting reactant when 5.00 g of H2
and 10.0 g of O2
react and form water?

Answer 1

To determine the limiting reactant in the reaction between hydrogen gas (H₂) and oxygen gas (O₂) to form water (H₂O), we can follow these steps:

1. **Write the balanced chemical equation** for the reaction:
\[
2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(l)
\]

2. **Calculate the molar masses**:
- Molar mass of H₂ = 2.02 g/mol
- Molar mass of O₂ = 32.00 g/mol

3. **Convert the masses of reactants to moles**:
- For H₂:
\[
\text{Moles of H}_2 = \frac{5.00 \text{ g}}{2.02 \text{ g/mol}} \approx 2.48 \text{ moles H}_2
\]
- For O₂:
\[
\text{Moles of O}_2 = \frac{10.0 \text{ g}}{32.00 \text{ g/mol}} \approx 0.3125 \text{ moles O}_2
\]

4. **Use the stoichiometry of the balanced equation** to find the ratio of reactants needed:
According to the balanced equation, 2 moles of H₂ react with 1 mole of O₂. Therefore, for 0.3125 moles of O₂:
- Required moles of H₂:
\[
0.3125 \text{ moles O}_2 \times \frac{2 \text{ moles H}_2}{1 \text{ mole O}_2} = 0.625 \text{ moles H}_2
\]

5. **Compare the available moles of H₂** with the required moles:
- We have 2.48 moles of H₂ available, but only 0.625 moles are needed to react with the available O₂.

Since we have more than enough H₂ to react with the available O₂, the limiting reactant must be **O₂**.

Thus, the limiting reactant in this reaction is **O₂**.