Question

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

experiment, 13.19 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction.

Answer 1

Approximately
`81.84\%`.

Step-by-step explanation:

Balanced equation for this reaction:

`{\rm Na_(2)CO_(3)}\, (aq) + {\rm CaCl_(2)} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_(3)}\, (s)`.

Look up the relative atomic mass of elements in the limiting reactant,
`\rm CaCl_(2)`, as well as those in the product of interest,
`\rm CaCO_(3)`:

• 
  `\rm Ca`:
  `40.078`.
• 
  `\rm Cl`:
  `35.45`.
• 
  `\rm C`:
  `12.011`.
• 
  `\rm O`:
  `15.999`.

Calculate the formula mass for both the limiting reactant and the product of interest:

`\begin{aligned}& M({\rm CaCl_(2)}) \\ &= (40.078 + 2 * 35.45)\; {\rm g \cdot mol^(-1)} \\ &= 110.978\; \rm g \cdot mol^(-1)\end{aligned}`.

`\begin{aligned}& M({\rm CaCO_(3)}) \\ &= (40.078 + 12.011 + 3 * 15.999)\; {\rm g \cdot mol^(-1)} \\ &= 100.086\; \rm g \cdot mol^(-1)\end{aligned}`.

Calculate the quantity of the limiting reactant (
`\rm CaCl_(2)`) available to this reaction:

`\begin{aligned}n({\rm CaCl_(2)) &= \frac{m({\rm {CaCl_(2)})}}{M({\rm CaCl_(2)})} \\ &= (17.87\; \rm g)/(110.978\; \rm g \cdot mol^(-1)) \\ &\approx 0.161023\; \rm mol \end{aligned}`.

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant (
`\rm CaCl_(2)`) and the product (
`{\rm CaCO_(3)}`) are both
`1`. Thus:

`\displaystyle \frac{n({\rm CaCO_(3)})}{n({\rm CaCl_(2)})} = 1`.

In other words, for every
`1\; \rm mol` of
`\rm CaCl_(2)` formula units that are consumed,
`1\; \rm mol\!` of
`\rm CaCO_(3)` formula units would (in theory) be produced. Thus, calculate the theoretical yield of
`\rm CaCO_(3)\!` in this experiment:

`\begin{aligned} & n(\text{${\rm CaCO_(3)}$, theoretical}) \\ =\; & n({\rm CaCl_(2)}) \cdot \frac{n({\rm CaCO_(3)})}{n({\rm CaCl_(2)})} \\ \approx \; & 0.161023\; {\rm mol} * 1 \\ =\; & 0.161023\; \rm mol\end{aligned}`.

Calculate the theoretical yield of this experiment in terms of the mass of
`\rm CaCO_(3)` expected to be produced:

`\begin{aligned} & m(\text{${\rm CaCO_(3)}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_(3)}$, theoretical}) \cdot M(({\rm CaCO_(3)}) \\ \approx \; & 0.161023\; {\rm mol} * 100.086\; {\rm g \cdot mol^(-1)} \\ \approx \; & 16.1161\; \rm g \end{aligned}`.

Given that the actual yield in this question (in terms of the mass of
`\rm CaCO_(3)`) is
`13.19\; \rm g`, calculate the percentage yield of this experiment:

`\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} * 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} * 100\% \\ \approx \; & 81.84\%\end{aligned}`.