Question

The zirconium oxalate \(\mathrm{K}_{2} \mathrm{Zr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right) \cdot \mathrm{H}_{2} \mathrm{O}\) was synthesized by mixing \(1.68 \mathrm{~g}\) of \(\mathrm{ZrOCl}_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) with \(5.20 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) and an excess of aqueous \(\mathrm{KOH}\). After \(2 \mathrm{months}\) \(1.25 \mathrm{~g}\) of crystalline product was obtained, along with aqueous \(\mathrm{KCl}\) and water. Calculate the percent yield.

Short answer

The percent yield is 42.23%.

Step-by-step solution

- Identify the chemical reaction

Determine the chemical reaction that takes place. The reaction involves the synthesis of zirconium oxalate using \(\text{ZrOCl}_{}_{2} \cdot 8 \text{H}_{}_{2} \text{O}\) and oxalic acid \(\text{H}_{}_{2} \text{C}_{}_{2} \text{O}_{}_{4} \cdot 2 \text{H}_{}_{2} \text{O}\) with excess \(\text{KOH}\). The reaction produces zirconium oxalate \(\text{K}_{2} \text{Zr}\text{(C}_{2}\text{O}_{4}\text{)}_{3}\text{(H}_{2}\text{C}_{2}\text{O}_{4}\text{)} \cdot \text{H}_{2} \text{O}\) and by-products.

- Calculate moles of reactants

Calculate the moles of each reactant used. For \(\text{ZrOCl}_{2} \cdot 8 \text{H}_{2} \text{O}\):\[ \text{Molar mass} = 322.24 \text{g/mol} \Rightarrow \text{Moles of} \text{ZrOCl}_{2} \cdot 8 \text{H}_{2} \text{O} = \frac{1.68 \text{g}}{322.24 \text{g/mol}} = 0.00521 \text{mol}. \]For \(\text{H}_{2} \text{C}_{2} \text{O}_{4} \cdot 2 \text{H}_{2} \text{O}\):\[ \text{Molar mass} = 126.07 \text{g/mol} \Rightarrow \text{Moles of} \text{H}_{2} \text{C}_{2} \text{O}_{4} \cdot 2 \text{H}_{2} \text{O} = \frac{5.20 \text{g}}{126.07 \text{g/mol}} = 0.0413 \text{mol}. \]

- Determine limiting reactant

Using stoichiometry, determine the limiting reactant. The balanced equation suggests a 1:3 ratio.With 0.00521 mol of \(\text{ZrOCl}_{}_{2} \cdot 8 \text{H}_{}_{2} \text{O}\):\[ 0.00521 \text{mol} \times 3 = 0.0156 \text{mol}. \]Given 0.0413 mol of \(\text{H}_{2} \text{C}_{2} \text{O}_{4} \cdot 2 \text{H}_{2} \text{O}\), \(\text{ZrOCl}_{}_{2} \cdot 8 \text{H}_{2} \text{O}\) is the limiting reactant.

- Calculate theoretical yield

Using moles of the limiting reactant to find the amount of the theoretical product formed.\Using the molar mass calculation: \(\text{K}_{}_{2} \text{Zr}\text{(C}_{}_{2} \text{O}_{}_{4}\text{)}_{3}\text{(H}_{}_{2} \text{C}_{}_{2} \text{O}_{}_{4} \cdot \text{H}_{}_{2} \text{O}\) has molar mass of 568.41 \text{g/mol}.\Using 0.00521 mol of \(\text{ZrOCl}_{}_{2} \text{cdot 8} \text{H}_{2} \text{O}\)\[ \text{Theoretical yield} = 0.00521 \text{mol} \times 568.41 \text{g/mol} = 2.96 \text{g}. \]

- Calculate percent yield

Using the formula for percent yield: \[\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100\%\]\Given actual yield of the product \(1.25 \text{g}\):\[ \text{Percent yield} = \frac{1.25 \text{g}}{2.96 \text{g}} \times 100\% = 42.23\%.\]

Key concepts

limiting reactant

Understanding the concept of a limiting reactant is crucial in stoichiometry. The limiting reactant is the substance that is totally consumed first in a chemical reaction. This reactant limits the amount of products formed and determines when the reaction will stop. To find the limiting reactant, you need to start with a balanced chemical equation.

Let's see this in the context of our example. Here, we mixed \( \text{ZrOCl}_{2} \cdot 8 \text{H}_{2} \text{O} \) and ::math2:: + hydroxyaloxysilanes:xxx:e with excess KOH. To find which reactant is the limiting one, we must calculate the moles of each reactant. From the formula above, we have 0.00521 mol of ::textdata:: using stoichiometry from the balanced equation, we realize that :textst data: mol of ZrOCl₂·8H₂O would need 3 times as many moles of H₂C₂O₄·2H₂O.

Comparing these values, we can determine that ::textdata:: is the limiting reactant because it will run out first, preventing more product from being formed.

theoretical yield

Theoretical yield refers to the maximum amount of product that can be produced in a chemical reaction if all the limiting reactant completely reacts. It’s calculated using the balanced chemical equation and stoichiometry.

To determine the theoretical yield, we'll need the moles of the limiting reactant and the molar mass of the desired product. In our example, the moles of ::textdata:: is 0.00521 mol, and molar mass is 568.41 g/mol.

By multiplying these together, we find:
\[ \text{Theoretical Yield} = 0.00521 \ mol \times 568.41 \ g/mol \approx 2.96 \ g \]

This is the maximum amount of zirconium oxalate we could theoretically produce.

stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It helps to predict the amounts of substances consumed and produced.

First, ensure you have a balanced chemical equation. Then, convert the quantities of your reactants into moles. This allows you to compare them using the mole ratios from the balanced equation.

In our problem, the balanced chemical equation would show a 1:3 ratio between \( \text{ZrOCl}_{2} \) and \( \text{H}_{2} \), respectively. Using this ratio, we can work out the moles needed and then apply it to find which reactant will limit the production, how much of the product can be theoretically produced, and compare it with actual yields.

Stoichiometry provides the foundational understanding needed to carry out these calculations accurately, showing you exactly how much of each substance reacts and forms.