Respuesta :
Answer:
Approximately [tex]81.84\%[/tex].
Explanation:
Balanced equation for this reaction:
[tex]{\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)[/tex].
Look up the relative atomic mass of elements in the limiting reactant, [tex]\rm CaCl_{2}[/tex], as well as those in the product of interest, [tex]\rm CaCO_{3}[/tex]:
- [tex]\rm Ca[/tex]: [tex]40.078[/tex].
- [tex]\rm Cl[/tex]: [tex]35.45[/tex].
- [tex]\rm C[/tex]: [tex]12.011[/tex].
- [tex]\rm O[/tex]: [tex]15.999[/tex].
Calculate the formula mass for both the limiting reactant and the product of interest:
[tex]\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
[tex]\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the quantity of the limiting reactant ([tex]\rm CaCl_{2}[/tex]) available to this reaction:
[tex]\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}[/tex].
Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ([tex]\rm CaCl_{2}[/tex]) and the product ([tex]{\rm CaCO_{3}}[/tex]) are both [tex]1[/tex]. Thus:
[tex]\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1[/tex].
In other words, for every [tex]1\; \rm mol[/tex] of [tex]\rm CaCl_{2}[/tex] formula units that are consumed, [tex]1\; \rm mol\![/tex] of [tex]\rm CaCO_{3}[/tex] formula units would (in theory) be produced. Thus, calculate the theoretical yield of [tex]\rm CaCO_{3}\![/tex] in this experiment:
[tex]\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} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}[/tex].
Calculate the theoretical yield of this experiment in terms of the mass of [tex]\rm CaCO_{3}[/tex] expected to be produced:
[tex]\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}[/tex].
Given that the actual yield in this question (in terms of the mass of [tex]\rm CaCO_{3}[/tex]) is [tex]13.19\; \rm g[/tex], calculate the percentage yield of this experiment:
[tex]\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}[/tex].
