Question

consider this hypothetical reaction: a + 3b → c + 3d. assume that the equation is balanced. in which of the reactant mole ratios given will reactant b be the limiting reagent? a/5b a/b a/4b 2a/7b

Explanation:

Step1: Recall limiting reagent concept

The balanced reaction is \( A + 3B
ightarrow C + 3D \). The stoichiometric mole ratio of \( A \) to \( B \) is \( \frac{n_A}{n_B}=\frac{1}{3}\approx0.333 \). For \( B \) to be the limiting reagent, the actual mole ratio \( \frac{n_A}{n_B} \) must be greater than the stoichiometric ratio (because more \( A \) relative to \( B \) would make \( B \) run out first).

Step2: Calculate each option's ratio

• Option A/5B: \( \frac{1}{5} = 0.2 \) (less than 0.333, so \( A \) is limiting)
• Option A/B: \( \frac{1}{1}=1 \) (greater than 0.333)
• Option A/4B: \( \frac{1}{4} = 0.25 \) (less than 0.333, \( A \) limiting)
• Option 2A/7B: \( \frac{2}{7}\approx0.286 \) (less than 0.333, \( A \) limiting) Wait, no, wait. Wait, the stoichiometric ratio is \( A:B = 1:3 \), so \( \frac{n_A}{n_B} \) stoichiometric is \( \frac{1}{3}\approx0.333 \). For \( B \) to be limiting, we need \( \frac{n_A}{n_B}>\frac{1}{3} \), because if we have more \( A \) per \( B \) than the stoichiometric ratio, \( B \) will be used up first. Wait, let's re - express:

From the reaction \( A + 3B
ightarrow C + 3D \), 1 mole of \( A \) reacts with 3 moles of \( B \). So the mole ratio of \( A \) to \( B \) required for complete reaction is \( \frac{n_A}{n_B}=\frac{1}{3}\approx0.333 \).

If the actual \( \frac{n_A}{n_B}>\frac{1}{3} \), it means we have more \( A \) relative to \( B \) than needed. So \( B \) will be the limiting reagent (because there is not enough \( B \) to react with all the \( A \)).

Now let's calculate each ratio:

• \( \frac{A}{5B}=\frac{1}{5} = 0.2<0.333 \): So we have less \( A \) per \( B \) than needed. So \( A \) is limiting.
• \( \frac{A}{B}=1>0.333 \): We have more \( A \) per \( B \) than needed. So \( B \) is limiting.
• \( \frac{A}{4B}=\frac{1}{4}=0.25 < 0.333\): \( A \) is limiting.
• \( \frac{2A}{7B}=\frac{2}{7}\approx0.286<0.333 \): \( A \) is limiting.

Ah, I made a mistake earlier. So the ratio \( \frac{A}{B} \) (1) is greater than \( \frac{1}{3}\approx0.333 \), so in this case, \( B \) is the limiting reagent.

Answer:

B. A/B