Question

question 12
1 pts
consider this hypothetical reaction: a + 3b \longrightarrow c + 3d. assume that the equation is balanced. in which of the reactant mole ratios given will reactant b be the limiting reagent?
\bigcirc a/b
\bigcirc a/5b
\bigcirc a/4b
\bigcirc 2a/7b

Explanation:

Step1: Determine the stoichiometric ratio of A and B

From the balanced equation \(A + 3B
ightarrow C + 3D\), the stoichiometric ratio of \(A\) to \(B\) is \(n(A)/n(B)=\frac{1}{3}\)

Step2: Analyze each option

• For option \(A/B\): The ratio of \(n(A)/n(B) = 1/1\). If we assume \(n(A)=1\) mole and \(n(B) = 1\) mole. Using the mole - ratio from the balanced equation \(A+3B

ightarrow C + 3D\), for \(1\) mole of \(A\), we need \(3\) moles of \(B\). But we have only \(1\) mole of \(B\). So \(B\) is the limiting reagent.

• For option \(A/5B\): The ratio \(n(A)/n(B)=\frac{1}{5}\). If \(n(A) = 1\) mole, we need \(3\) moles of \(B\) (from the balanced equation). We have \(5\) moles of \(B\). So \(A\) is the limiting reagent.
• For option \(A/4B\): The ratio \(n(A)/n(B)=\frac{1}{4}\). If \(n(A)=1\) mole, we need \(3\) moles of \(B\). We have \(4\) moles of \(B\). So \(A\) is the limiting reagent.
• For option \(2A/7B\): The ratio \(n(A)/n(B)=\frac{2}{7}\approx0.29\). From the balanced equation \(n(A)/n(B)=\frac{1}{3}\approx0.33\). Cross - multiply: \(2\times3 = 6\) and \(7\times1=7\). Since \(6<7\), \(A\) is the limiting reagent

Answer:

A. \(A/B\)