Q3.127CP - Chemistry: Molecular Nature Of Matter And Change

Question

The zirconium oxalate K2Zr(C2O4)3(H2C2O). H2O was synthesized by mixing 1.68 g of ZrOCl2 .8H2O with 5.20 g of H2C2O4.2H2O and an excess of aqueous KOH. After 2 months, 1.25 g of crystalline product was obtained, as well as aqueous KCl and water. Calculate the percent yield.

Step-by-Step Solution

Answer

Answer

The percent yield is 44.3%

Step 1: Writing the balanced chemical equation

$$\begin{gathered} {\mathrm{ZrOCl}}_2 · 8{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}+4{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4 · 2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)} +4{\mathrm{KOH}}_{\left(\mathrm{aq}\right)} \\ \to {\mathrm{K}}_2\mathrm{Zr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\left({\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\right) · {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)} + 2{\mathrm{KCl}}_{\left(\mathrm{aq}\right)} + 20 {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{aq}\right)} \end{gathered}$$

Step 2: Determine the limiting reactant

The limiting reactant is the one which produces less product. Here two reactant is provided. To determine which reactant is limiting reactant, you have to calculate the amount of product formed from each reactant. This can be done by keeping the other reactant in excess. KOH is in excess so the moles of product formed from KOH is not to be calculated. 

Assuming H2C2O4.2H2 O is limiting.

Finding the moles of product formed from the amount of ZrOCl2 .8H2O

$$\begin{gathered} \left(1.68 g ZrOC{l}_2 · 8H_{2}O\right)\left(\frac{1 mol ZrOC{l}_2 · 8H_{2}O}{322.25 g ZrO{l}_2 · 8H_{2}O}\right)\left(\frac{1 mol of product}{1 mol ZrOC{l}_2 · 8H_{2}O}\right) \\ =0.005213 mol product \end{gathered}$$

Assuming ZrOCl2 .8H2O is limiting.

Finding the moles of product formed from the amount of H2C2O4.2H2 O

$$\begin{gathered} (5.20 g H_2{C}_2{O}_4 · 2H_{2}O)\left(\frac{1 mol H_2{C}_2{O}_4 · 2H_{2}O}{126.07 g H_2{C}_2{O}_4 · 2H_{2}O}\right)\left(\frac{1 mol of product}{1 mol H_2{C}_2{O}_4 \boxed{·} 2H_{2}O}\right) \\ =0.010311 mol product \end{gathered}$$

From the above calculation, ZrOCl2 .8H2O is the limiting reactant. And is use to calculate the theoretical yield.

$$\begin{gathered} \mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\left(g\right)\mathbf{ }\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathbf{=}\left(0.005213 mol product\right)\left(\frac{541.53 g product}{1 mol product}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{8231}\mathbf{ }\mathit{g}\mathbf{ }\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t} \end{gathered}$$

Step 3: Calculating the percentage yield

The amount of product that you actually obtain is the actual yield. The percent yield is the actual yield expressed as a percentage of the theoretical yield.

$$\begin{gathered} \mathbf{\%}\mathbf{ }\mathit{y}\mathit{i}\mathit{e}\mathit{l}\mathit{d}\mathbf{=}\frac{\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{u}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{y}\mathbf{i}\mathbf{e}\mathbf{l}\mathbf{d}}{\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{o}\mathbf{r}\mathbf{e}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{a}\mathbf{l}\mathbf{ }\mathbf{y}\mathbf{i}\mathbf{e}\mathbf{l}\mathbf{d}}\mathbf{\times }\mathbf{100} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{ }\mathbf{g}}{\mathbf{2}\mathbf{.}\mathbf{8231}\mathbf{ }\mathbf{g}}\mathbf{\times }\mathbf{100} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{44}\mathbf{.}\mathbf{3}\mathbf{\%} \end{gathered}$$