Question

a. Identify the limiting reagent when 15.0 g of H2 reacts with 48.0 g of O2 to produce water.
b. What is the mass of water produced?

Answer 1

a. Identify the limiting reagent when 15.0 g of H2 reacts with 48.0 g of O2 to produce water.

b. What is the mass of water produced?

**Solution:**

We are given:

$$m(H_2) = 15 \, g$$ $$m(O_2) = 48 \, g$$

Reaction equation:

$$2H_2 + O_2 = 2H_2O$$

Amount of substance of $H_2$:

$$v(H_2) = \frac{m(H_2)}{Mr(H_2)} = \frac{15}{2 \cdot 1} = 7.5 \, \text{mol}$$

Amount of substance of $O_2$:

$$v(O_2) = \frac{m(O_2)}{Mr(O_2)} = \frac{48}{16 \cdot 2} = 1.5 \, \text{mol}$$

According to the reaction equation:

$$\frac{v(H_2)}{v(O_2)} = \frac{2}{1}$$

Obviously, oxygen $O_2$ is the limiting reagent. Only 3 mol of $H_2$ are required to react with 1.5 mol of $O_2$.

According to the reaction equation:

$$\frac{v(O_2)}{v(H_2O)} = \frac{1}{2}$$

Thus:

$$v(H_2O) = 1.5 \cdot 2 = 3 \, \text{mol}$$ $$m(H_2O) = v(H_2O) \cdot Mr(H_2O) = 3 \cdot 18 = 54 \, g$$

**Answer:**

a. $O_2$

b. $54 \, g$

**References:**

1. http://en.wikipedia.org/wiki/Molecular_mass