# Limiting Reagent Question?

May 23, 2017

$\text{LiOH}$ is limiting, and

$2.00 {\text{mol Li"_2"CO}}_{3}$ forms.

#### Explanation:

We can start by writing the chemical equation for this reaction:

$2 \text{LiOH(aq)" + "CO"_2"(g)" rarr "Li"_2"CO"_3 "(aq)" + "H"_2"O(l)}$

A simple way to find the limiting reagent is to calculate the number of moles of ${\text{Li"_2"CO}}_{3}$ that can form from each $4.00 \text{mol LiOH}$ and $3.20 {\text{mol CO}}_{2}$. Whichever reagent produces the lesser amount of ${\text{Li"_2"CO}}_{3}$ is limiting;

4.00cancel("mol LiOH")((1"mol Li"_2"CO"_3)/(2cancel("mol LiOH"))) = color(red)(2.00"mol Li"_2"CO"_3

3.20cancel("mol CO"_2)((1"mol Li"_2"CO"_3)/(1cancel("mol CO"_2))) = color(blue)(3.20"mol Li"_2"CO"_3

We can see that the $4.00 \text{mol LiOH}$ produces less ${\text{Li"_2"CO}}_{3}$, so the $\text{LiOH}$ is the limiting reagent, and $\textcolor{red}{2.00 {\text{mol Li"_2"CO}}_{3}}$ will form.