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University of Maine

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Problem 112

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 months, 1.25 $\mathrm{g}$ of crystalline product was obtained, along with aqueous $\mathrm{KCl}$ and water. Calculate the percent yield.

Answer

percent yield=n(exp)/n(theor) $^{*} 100=44,39$ percent

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## Discussion

## Video Transcript

for the following reaction. The first step is to be sure to balance it before looking at calculating the percent yield. When this equation is balanced, the coefficients are one for four, and then the products are one to and 20. It's also helpful before you start to find the molar mass of the compounds involved in the reaction. We find this by adding up the individual molar masses, including the hydrates or the waters using the periodic table. So for the first compound, the molar mass is 322 0.24 grams per mole. My second reacted has a molar mass of 126 0.7 grams per mole. And finally, the product that we form here has a molar mass of 541 0.66 grams per mole. So the question tells us that we start off with an initial amount of both reactive. We have 1.68 grams of the zirconium reactant, and we have five 0.20 grams of the actually reacted, and we know that were produced 1.25 grams of the product to find the percent yield we need to figure out what the theoretical yield is or how much is actually produced, so we need to solve for the theoretical yield or the calculated amount. So to do that, we have both reactors Ingrams, and we change two moles using the molar Mass. And then we have moles of the compound. You changed two moles of the product. We do that by using the balanced equations, the mole ratio from the balanced equation. And then finally, we can change two grams the product using the molar mass of the product. Because we're given both reactions, we need to do this steps for each one and look for which reactant produces the smaller amount. This will be our theoretical yield, so the smaller amount produced is our theoretical yield. Let's start with zirconium complex 1.6 grams, 68 grams to change two moles by dividing by the molar Mass and then, in looking at the balanced equation, receive that it the 1 to 1 ratio between the product and the zirconium, or one mole of product for every one mole of zero r o c. L 2.8 h 20 And then we change from grams, two moles of product multiplying by the molar mass or 541 0.66 grams for every mole of product. Or we produced 2.83 grams. Similarly looking at the second reacted H two C two 04 not to H 20 when we started out with five 0.20 grams, we change it to moles using the molar Mass. 126 0.7 grams. And then, in this case, when we look at the balanced equation, there are formals of reactive for every one mole of product, and then we multiply by 5 41.66 grams per mole where were produced 5.59 grands. This number is the smaller. So that's the theoretical yield to our percent yield is how much we actually produced, which is 1.25 grams, divided by what we should have produced 2.83 grams times 100 or 44 percent

## Recommended Questions

Van Arkel Process Pure zirconium is obtained using the two-step $\mathrm{Van}$ Arkel process. In the first step, impure zirconium and iodine are heated to produce zirconium iodide (ZrI_). In the second step, ZrI_ is decomposed to produce pure zirconium.

$$\mathrm{ZrI}_{4}(\mathrm{s}) \rightarrow \mathrm{Zr}(\mathrm{s})+2 \mathrm{I}_{2}(\mathrm{g})$$

When 8.50 g of carbon monoxide reacts with an excess of hydrogen, 8.52 g of methanol is collected. Complete Table 11.4, and calculate the percent yield.

Van Arkel Process Pure zirconium is obtained using

the two-step Van Arkel process. In the first step, impure

zirconium and iodine are heated to produce zirconium

iodide $\left(Z r I_{4}\right) .$ In the second step, ZrI $_{4}$ is decomposed to

produce pure zirconium.

\begin{equation}

\mathrm{ZrI}_{4}(\mathrm{s}) \rightarrow \mathrm{Zr}(\mathrm{s})+2 \mathrm{I}_{2}(\mathrm{g})

\end{equation}

Determine the percent yield of zirconium if 45.0 $\mathrm{g}$ of

$\mathrm{ZrI}_{4}$ is decomposed and 5.00 $\mathrm{g}$ of pure Zr is obtained.

Uranium can be isolated from its ores by dissolving it as UO $_{2}\left(\mathrm{NO}_{3}\right)_{2},$ then separating it as solid

$\mathrm{UO}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right) \cdot 3 \mathrm{H}_{2} \mathrm{O}$ . Addition of 0.4031 $\mathrm{g}$ of sodium oxalate, $\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4},$ to a soluting 1.481 $\mathrm{g}$ of uranyl nitrate, $\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2},$ yields 1.073 $\mathrm{g}$ of solid $\mathrm{UO}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right) \cdot 3 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{UO}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right) \cdot 3 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NaNO}_{3}$

Determine the limiting reactant and the percent yield of this reaction.

Sodium oxalate, Na2C2O4, in solution is oxidized to CO2(g) by MnO4-, which is reduced to Mn2+. A 50.1-mL volume of a solution of MnO4- is required to titrate a 0.339 g sample of sodium oxalate. This solution of MnO4- is used to analyze uranium-containing samples. A 4.62-g sample of a uranium-containing material requires 32.5 mL of the solution for titration. The oxidation of the uranium can be represented by the change UO2+-UO22+. Calculate the percentage of uranium in the sample.

Titanium(IV) oxide ( $\mathrm{Ti} \mathrm{O}_{2}$ ) is a white substance produced by the action of sulfuric acid on the mineral ilmenite (FeTiO_):

$$

\mathrm{FeTiO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{TiO}_{2}+\mathrm{FeSO}_{4}+\mathrm{H}_{2} \mathrm{O}

$$

Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process, $8.00 \times 10^{3} \mathrm{kg}$ of $\mathrm{FeTiO}_{3}$ yielded $3.67 \times 10^{3} \mathrm{kg}$ of

TiO $_{2}$. What is the percent yield of the reaction?

Upon heating, calcium carbonate $\left(\mathrm{CaCO}_{3}\right)$ decomposes to calcium oxide $(\mathrm{CaO})$ and carbon dioxide $\left(\mathrm{CO}_{2}\right)$

a. Determine the theoretical yield of $\mathrm{CO}_{2}$ if 235.0 $\mathrm{g}$ of $\mathrm{CaCO}_{3}$ is heated.

b. What is the percent yield of $\mathrm{CO}_{2}$ if 97.5 $\mathrm{g}$ of $\mathrm{CO}_{2}$ is

collected?

Titanium(IV) oxide (TiO_) is a white substance produced by the action of sulfuric acid on the mineral ilmenite (FeTiO $_{3}$ ):

$$\mathrm{FeTiO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{TiO}_{2}+\mathrm{FeSO}_{4}+\mathrm{H}_{2} \mathrm{O}$$

Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process $8.00 \times 10^{3} \mathrm{kg}$ of $\mathrm{FeTiO}_{3}$ yielded $3.67 \times 10^{3} \mathrm{kg}$ of $\mathrm{TiO}_{2} .$ What is the percent yield of the reaction?

You mix a 25.0 mL sample of a 1.20 M potassium chloride solution with 15.0 mL of a 0.900 M barium nitrate solution, and this precipitation reaction occurs:

2 KCl(aq) + Ba(NO3)2(aq)-BaCl2(s) + 2 KNO3(aq)

You collect and dry the solid BaCl2 and find it has a mass of 2.45 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Zirconia $\left(\mathrm{ZrO}_{2}\right),$ an unusually tough oxide ceramic, has been used to make very sharp table knives. Write a balanced equation for the hydrolysis of zirconium isopropoxide in the sol–gel method for making zirconia powders. The isopropoxide ligand is the anion of isopropyl alcohol, $\mathrm{HOCH}\left(\mathrm{CH}_{3}\right)_{2}$

A 25.0 -mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 $\mathrm{mL}$ of a 0.900 $\mathrm{M}$ lead(II) nitrate solution and this precipitation reaction occurs:

$$2 \mathrm{KCl}(a q)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{PbCl}_{2}(s)+2 \mathrm{KNO}_{3}(a q)$$

The solid $\mathrm{PbCl}_{2}$ is collected, dried, and found to have a mass of 2.45 $\mathrm{g} .$ Determine the limiting reactant, the theoretical yield, and the percent yield.