# Help with these questions about limiting reactants and yield?

## 1) Given the reaction: $3 X + 4 Y \to 5 Z$ If after reacting $\text{2.1 mols}$ of $X$ with $\text{1.3 mols}$ of $Y$, the percent yield is 50%, and the actual yield is $\text{75.0 g}$ of $Z$, what is the molar mass of $Z$? 2) What is the theoretical yield of ${\text{SiCl}}_{4}$ if $\text{2 g}$ of $\text{Si}$ react with ${\text{1 g Cl}}_{2}$? ${\text{Si"(s) + 2"Cl"_2(g) -> "SiCl}}_{4} \left(s\right)$

Oct 8, 2016

1) Since you were asked to find the molar mass, which is in $\boldsymbol{\text{g/mol}}$, you must obtain the product mass in $\boldsymbol{\text{g}}$ and the product quantity in $\boldsymbol{\text{mol}}$s.

Note that these (mass + mols) must be based on theoretical yield, and not actual yield (because in actuality you may have lost reactant while the reaction went on, which reduces the true yield and creates errors in your calculations later on).

• You were given percent yield and actual yield, which allows you to find the theoretical mass yield.
• You were given the balanced chemical equation, which you can use to find the theoretical product quantity in $\text{mol}$s.

So, we do the following calculations.

$\text{% Yield" = "Actual Yield"/"Theoretical Yield}$

=> color(green)("Theoretical Yield") = "Actual Yield"/"% Yield"

= "75.0 g Z"/0.50 = color(green)("150.0 g Z")

The $\text{mol}$s of $Z$ are then based on the limiting reactant. Based on the chemical reaction, you have a $3 : 4$ ratio of $X : Y$, but you only have a $2.1 : 1.3$ ratio of $X : Y$, or a $3 : 1.86$ ratio.

That means you have too little $Y$, and $Y$ is your limiting reactant, limiting the amount of product you can make.

Thus, your $\text{mol}$s of $Z$ are:

$\textcolor{g r e e n}{\text{mols Z") = "1.3 mols Y" xx "5 mols Z"/"4 mols Y" = color(green)("1.625 mols Z}}$

(You should find that using $X$ gives you $\text{3.5 mols Z}$, which is not theoretically possible because $Y$ ran out already before the remaining expected $3.5 - 1.625 = \text{1.875 mols}$ of $Z$ could be produced.)

$\textcolor{b l u e}{{M}_{m , Z}} = \text{150.0 g Z"/"1.625 mols Z}$

$=$ $\textcolor{b l u e}{\text{92.31 g/mol}}$

2) This is a similar problem with limiting reactants, but this is simpler.

In this case, try picking each reactant and converting to $\text{mol}$s of it, then $\text{mol}$s of product, and convincing yourself which one yields less product and is thus the limiting reactant.

That means assume one reactant is the limiting reactant, and check to see which one yields less product. In general:

"? g SiCl"_4 = "? g A" xx "mols A"/"? g A" xx ("1 mol SiCl"_4)/"? mol A" xx ("? g SiCl"_4)/("mol SiCl"_4)

"? g SiCl"_4 = "? g B" xx "mols B"/"? g B" xx ("1 mol SiCl"_4)/"? mol B" xx ("? g SiCl"_4)/("mol SiCl"_4)

Then, using the limiting reactant, you can find the theoretical yield of ${\text{SiCl}}_{4}$, which is your goal.