# "13.1 g CaO" react with "38.9 g CO"_2". What is the percent yield if "16.7 g CaCO"_3" are produced?

Mar 16, 2017

The limiting reactant is $\text{CaO}$. The percent yield is "71.4%.

#### Explanation:

Balanced equation.

${\text{CaO + CO}}_{2}$$\rightarrow$$\text{CaCO"_3}$

This is a limiting reactant question. We need to determine the mass of $\text{CaCO"_3}$ that can be produced by the given masses of $\text{CaO}$ and $\text{CO"_2}$. The one that produces the least mass of $\text{CaCO"_3}$ is the limiting reactant, and the mass of $\text{CaCO"_3}$ produced by that reactant is the theoretical yield. Once we have the mass of $\text{CaCO"_3}$ produced, we can determine percent yield.

The process will include the following steps:

$\text{mass reactant}$$\rightarrow$$\text{mol reactant}$$\rightarrow$$\text{mol product}$$\rightarrow$$\text{mass product}$

Since there are no coefficients, we know that all of the mole ratios between the compounds are 1:1.

We need the molar masses for all compounds.
$\text{CaO} :$$\text{56.077 g/mol}$
$\text{CO"_2} :$$\text{44.009 g/mol}$
$\text{CaCO"_3} :$$\text{100.086 g/mol}$
https://www.ncbi.nlm.nih.gov/pccompound?term=CaO
https://www.ncbi.nlm.nih.gov/pccompound/?term=CO2
https://www.ncbi.nlm.nih.gov/pccompound/?term=CaCO3

We need to determine the moles of each reactant by dividing the given masses by their molar masses.

(13.1cancel("g") "CaO")/(56.077cancel"g"/"mol")="0.23361 mol CaO"

(38.9cancel("g") "CO"_2)/(44.009cancel("g")/"mol")="0.88391 mol CO"_2"

Convert mol $\text{CaO}$ to mol $\text{CaCO"_3}$ to mass $\text{CaCO"_3}$ by multiplying mol $\text{CaO}$ by the mol ratio that gives mol $\text{CaO}$ then multiply by the molar mass of $\text{CaCO"_3}$.

$0.23361 \cancel{\text{mol CO"xx(1cancel"mol CaCO"_3)/(1cancel"mol CO")xx100.086"g CaO"_3="23.4 g CaO"_3}}$

$\textcolor{b l u e}{\text{13.1 g CaO"_3}}$$\textcolor{b l u e}{\text{reacts with}}$ ${\textcolor{b l u e}{\text{CO}}}_{2}$$\textcolor{w h i t e}{.} \textcolor{b l u e}{\text{to produce}}$ color(blue)("23.4 g CaO"_3".

0.88391cancel("mol CO"_2)xx(1cancel"mol CaCO"_3)/(1cancel"mol CO"_2)xx(100.086"g CO"_2)/(1cancel"mol CaCO"_3)="88.5 g CaCO"_3"

$\textcolor{b l u e}{\text{38.9 g CO"_2}}$ $\textcolor{b l u e}{\text{react with}}$ $\textcolor{b l u e}{\text{CaO}}$$\textcolor{w h i t e}{.} \textcolor{b l u e}{\text{to produce}}$ $\textcolor{b l u e}{\text{88.5 g CaCO"_3}}$.

The limiting reagent is $\text{CaO}$, so the maximum amount of $\text{CaO"_3}$ that can be produced is $\text{24.4 g}$. This is the theoretical yield.

$\text{percent yield"=("actual yield")/("theoretical yield")xx100}$

$\text{percent yield"=("16.7 g CaCO"_3)/("23.4 g CaCO"_3)xx100="71.4%}$