Limiting Reagent Question?

1 Answer
May 23, 2017

"LiOH" is limiting, and

2.00 "mol Li"_2"CO"_3 forms.

Explanation:

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

2"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 "Li"_2"CO"_3 that can form from each 4.00"mol LiOH" and 3.20"mol CO"_2. Whichever reagent produces the lesser amount of "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"mol LiOH" produces less "Li"_2"CO"_3, so the "LiOH" is the limiting reagent, and color(red)(2.00"mol Li"_2"CO"_3) will form.