use the following information to answer the next three questions. note: in order to receive full marks, all…

use the following information to answer the next three questions. note: in order to receive full marks, all balanced equations, mole ratios, formulas and units must be shown. 19.3 g of chromium (ii) sulfate solution is reacted with 9.77 g of potassium hydroxide solution to produce a precipitate. written response 1. write the balanced chemical reaction including proper states. your answer: written response 2. calculate the maximum theoretical yield of the precipitate that is formed in this reaction by first finding the limiting reagent. show all work and record below. your answer: written response 3. if 8.85 g of the precipitate is actually produced, what is the percent yield and percent difference? (2 marks) what could account for the error in this lab? (1 mark) your answer:

use the following information to answer the next three questions. note: in order to receive full marks, all balanced equations, mole ratios, formulas and units must be shown. 19.3 g of chromium (ii) sulfate solution is reacted with 9.77 g of potassium hydroxide solution to produce a precipitate. written response 1. write the balanced chemical reaction including proper states. your answer: written response 2. calculate the maximum theoretical yield of the precipitate that is formed in this reaction by first finding the limiting reagent. show all work and record below. your answer: written response 3. if 8.85 g of the precipitate is actually produced, what is the percent yield and percent difference? (2 marks) what could account for the error in this lab? (1 mark) your answer:

Answer

Explanation:

Step1: Write the balanced chemical reaction

The reaction between chromium (II) sulfate ($CrSO_4$) and potassium hydroxide ($KOH$) is a double - displacement reaction. The products are chromium (II) hydroxide ($Cr(OH)_2$) precipitate and potassium sulfate ($K_2SO_4$). The balanced chemical equation with states is: $$CrSO_4(aq)+2KOH(aq)\rightarrow Cr(OH)_2(s)+K_2SO_4(aq)$$

Step2: Calculate the moles of reactants

The molar mass of $CrSO_4$ is $M_{CrSO_4}=52 + 32+4\times16=148\ g/mol$. The moles of $CrSO_4$, $n_{CrSO_4}=\frac{19.3\ g}{148\ g/mol}=0.1304\ mol$. The molar mass of $KOH$ is $M_{KOH}=39 + 16+1 = 56\ g/mol$. The moles of $KOH$, $n_{KOH}=\frac{9.77\ g}{56\ g/mol}=0.1745\ mol$.

Step3: Determine the limiting reagent

From the balanced equation, the mole ratio of $CrSO_4$ to $KOH$ is $1:2$. For $0.1304\ mol$ of $CrSO_4$, the moles of $KOH$ required is $2\times0.1304 = 0.2608\ mol$. But we have only $0.1745\ mol$ of $KOH$. For $0.1745\ mol$ of $KOH$, the moles of $CrSO_4$ required is $\frac{0.1745}{2}=0.08725\ mol$. Since we have more $CrSO_4$ than required, $KOH$ is the limiting reagent.

Step4: Calculate the theoretical yield of the precipitate

The mole ratio of $KOH$ to $Cr(OH)_2$ is $2:1$. The moles of $Cr(OH)2$ formed is $n{Cr(OH)_2}=\frac{0.1745\ mol}{2}=0.08725\ mol$. The molar mass of $Cr(OH)2$ is $M{Cr(OH)_2}=52+(16 + 1)\times2=86\ g/mol$. The theoretical yield of $Cr(OH)2$, $m{theoretical}=0.08725\ mol\times86\ g/mol = 7.4935\ g$.

Step5: Calculate the percent yield

The percent yield is given by $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100%$. Given actual yield $m_{actual}=8.85\ g$. $\text{Percent Yield}=\frac{8.85\ g}{7.4935\ g}\times100% = 118.1%$. The percent difference is $\text{Percent Difference}=\left|\frac{\text{Actual - Theoretical}}{\text{Theoretical}}\right|\times100%=\left|\frac{8.85 - 7.4935}{7.4935}\right|\times100%=18.1%$. Possible sources of error in the lab could be:

  1. Contamination of the reactants or products.
  2. Incomplete drying of the precipitate, leading to an over - estimation of the mass.
  3. Errors in weighing the reactants or products.

Answer:

  1. $CrSO_4(aq)+2KOH(aq)\rightarrow Cr(OH)_2(s)+K_2SO_4(aq)$
  2. The limiting reagent is $KOH$ and the theoretical yield of $Cr(OH)_2$ is $7.4935\ g$.
  3. Percent yield is $118.1%$, percent difference is $18.1%$. Possible sources of error include contamination, incomplete drying, and weighing errors.