Question

Use the balanced equation to answer the following questions.CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)A. What is the ratio of moles of CuSO4 to moles of Na2SO4B. What is the ratio of moles of CuSO4 to moles of NaOH_2. Determine the limiting reagent if 798.1 g CuSO4 reacts with 280.0 NaOH. Show all your calculations below. Justify your answer.3. Use the limiting reagent to determine how many grams of Cu(OH)2 should precipitate out in the reaction. Show your work4. If only 174.6 g of Cu(OH)2 precipitate were actually collected from the reaction, what would the percent yield be? Show your work

Answer 1

In this case, we have NaOH as the limiting agent. Therefore the reaction will be according to the number of moles of NaOH present. We have the following amount of NaOH:

The molecular weight (MW) of NaOH: 40.0 g/mol

`\begin{gathered} \text{NaOH mol: }\frac{\text{Mass of NaOH}}{\text{MWof NaOH}} \\ \text{NaOH mol: }\frac{\text{2}80.0\text{ g}}{\text{4}0.0\text{ g/mol}}=\text{ 7 mol of NaOH} \end{gathered}`

We have 7 mol of NaOH, so we will have 3.5 mol of Cu(OH)2. Because the ratio NaOH: Cu(OH)2 is 2:1. Now, we have to determine the grams of Cu(OH)2 using the molecular weight:

`\begin{gathered} \text{Mass of Cu(OH)2 }=\text{ MW Cu(OH)2 }*\text{ Mol of Cu(OH)2} \\ \text{Mass of Cu(OH)2 }=\text{ 97.6 g/mol }*\text{ 3}.5\text{ mol of Cu(OH)2} \\ \text{Mass of Cu(OH)2 }=\text{ 341.6 g of Cu(OH)2} \end{gathered}`

So, we will have 341.6 g of Cu(OH)2 precipitated after the reaction