The zirconium oxalate K2 Zr(C2 O4)3(H2 C2 O4) ·H2 O was synthesized by mixing 1.68 g of ZrOCl2

The zirconium oxalate K2 Zr(C2 O4)3(H2 C2 O4) ·H2 O was...

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.

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.

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Key Concepts

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It provides the basis for predicting how much product will form from given amounts of reactants by using balanced chemical equations, ensuring that the law of conservation of mass is satisfied.

Limiting Reagent

The limiting reagent is the reactant in a chemical reaction that is completely consumed first, thereby limiting the amount of product that can be formed. Identifying the limiting reagent is essential in stoichiometric calculations, as it determines the theoretical yield of the reaction.

Theoretical Yield

The theoretical yield is the maximum amount of product that could be produced from a given amount of reactants, as calculated from the balanced chemical equation under ideal conditions. It serves as a benchmark to compare with the actual yield observed in an experiment.

Percent Yield Calculation

Percent yield is a measure of the efficiency of a reaction, calculated by dividing the actual yield obtained from an experiment by the theoretical yield and then multiplying by 100. This percentage indicates how closely the experimental outcome approaches the ideal yield predicted by stoichiometry.

Transcript

00:01 For the following reaction, the first step is to be sure to balance it before looking at calculating the percent yield.

00:07 When this equation is balanced, the coefficients are 1, 4, 4, and then the products are 1, 2, and 20.

00:22 It's also helpful before we start to find the molar mass of the compounds involved in the reaction.

00:28 We find this by adding up the individual molar masses, including the hydrates or the waters, using the periodic table.

00:36 So for the first compound, the molar mass is 322 .24 grams per mole.

00:46 My second reactant has a molar mass of 126 .07 grams per mole.

00:55 And finally, the product that we form here has a molar mass of 524.

00:59 41 .66 grams per mole.

01:05 So the question tells us that we start off with an initial amount of both reactants.

01:12 We have 1 .68 grams of the zirconium reactant, and we have 5 .20 grams of the oxalate reactant.

01:25 And we know that we produce 1 .25 grams of the product.

01:31 To find the percent yield, we need to figure out what the theoretical yield is or how much is actually produced.

01:43 So we need to solve for the theoretical yield or the calculated amount.

01:51 So to do that, we have both reactants in grams and we change to moles using the molar mass.

02:07 And then we have moles of the compound, we change to moles of the product...

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