To determine the limiting reactant in this reaction between hydrogen ($\text{H}_2$) and oxygen ($\text{O}_2$) to form water ($\text{H}_2\text{O}$), we follow these steps:

1. Write the balanced chemical equation:

$2 \text{H}_2 (g) + \text{O}_2 (g) \rightarrow 2 \text{H}_2\text{O} (l)$

This shows that 2 moles of $\text{H}_2$ react with 1 mole of $\text{O}_2$.

2. Calculate the moles of each reactant:

   • The molar mass of $\text{H}_2$ is 2.02 g/mol.
   • The molar mass of $\text{O}_2$ is 32.00 g/mol.

$\text{Moles of } \text{H}_2 = \frac{5.00 \, \text{g}}{2.02 \, \text{g/mol}} = 2.475 \, \text{mol}$ $\text{Moles of } \text{O}_2 = \frac{10.0 \, \text{g}}{32.00 \, \text{g/mol}} = 0.3125 \, \text{mol}$

3. Compare the mole ratio with the balanced equation:

According to the balanced equation, 2 moles of $\text{H}_2$ react with 1 mole of $\text{O}_2$. Thus, we check how much of each reactant is required:

• For 0.3125 mol of $\text{O}_2$, the required amount of $\text{H}_2$ is: $\text{Required } \text{H}_2 = 0.3125 \, \text{mol} \times 2 = 0.625 \, \text{mol}$

Since you have 2.475 mol of $\text{H}_2$, which is much more than the 0.625 mol needed, $\text{O}_2$ is the limiting reactant.

4. Conclusion: The limiting reactant is oxygen ($\text{O}_2$).

Would you like more details, or do you have any further questions?

Related questions:

1. How do you determine the amount of product formed from the limiting reactant?
2. What happens to the excess reactant after the reaction?
3. How do you calculate the theoretical yield of water in this reaction?
4. What is the significance of the limiting reactant in industrial chemical reactions?
5. How would the reaction change if the masses of $\text{H}_2$ and $\text{O}_2$ were different?

Tip:

Always start with a balanced chemical equation before performing any stoichiometric calculations!