Question

which choice identifies the correct limiting reactant and correct reasoning?
2h₂ + o₂ → 2h₂o
0.40 g h₂ theoretically yields 0.20 mol h₂o.
1.8 g o₂ theoretically yields 0.11 mol h₂o.
h₂ because it has the higher theoretical yield
o₂ because it has the higher initial mass
h₂ because it has the lower initial mass
o₂ because it has the lower theoretical yield

Explanation:

To determine the limiting reactant, we compare the theoretical yields of the products from each reactant. The limiting reactant is the one that produces the least amount of product (has the lower theoretical yield) because it will be completely consumed first, limiting the amount of product formed.

• For $\ce{H_2}$: theoretical yield of $\ce{H_2O}$ is $0.20\ \text{mol}$.
• For $\ce{O_2}$: theoretical yield of $\ce{H_2O}$ is $0.11\ \text{mol}$.

Since $0.11\ \text{mol}$ (from $\ce{O_2}$) is less than $0.20\ \text{mol}$ (from $\ce{H_2}$), $\ce{O_2}$ is the limiting reactant because it has the lower theoretical yield.

Let’s analyze the other options:

• "H₂ because it has the higher theoretical yield": The limiting reactant is the one with the lower theoretical yield, not higher. Eliminate.
• "O₂ because it has the higher initial mass": Initial mass does not determine the limiting reactant—we need to compare moles or theoretical yields. $0.40\ \text{g}\ \ce{H_2}$ is less mass than $1.8\ \text{g}\ \ce{O_2}$, but mass alone isn’t the factor. Eliminate.
• "H₂ because it has the lower initial mass": Initial mass is irrelevant to identifying the limiting reactant (we use mole ratios or theoretical yields). Eliminate.

Answer:

O₂: because it has the lower theoretical yield