Use the balanced equation to answer the following questions.CuSO4(aq) + 2NaOH(aq) —->Cu(OH)2(s) + Na2SO4(aq)i. What is the ratio of moles of CuSO4 to moles of NaOH?ii. If 638.44 g CuSO4 reacts with 240.0 NaOH, which is the limiting reagent?iii. Using the limiting reagent to determine how many grams of Cu(OH)2 should precipitate out in the reaction.iv. If only 174.6g of Cu(OH)2 precipitate were actually collected from the reaction, what would the percent yield be?
Question
Use the balanced equation to answer the following questions.CuSO4(aq) + 2NaOH(aq) —->Cu(OH)2(s) + Na2SO4(aq)i. What is the ratio of moles of CuSO4 to moles of NaOH?ii. If 638.44 g CuSO4 reacts with 240.0 NaOH, which is the limiting reagent?iii. Using the limiting reagent to determine how many grams of Cu(OH)2 should precipitate out in the reaction.iv. If only 174.6g of Cu(OH)2 precipitate were actually collected from the reaction, what would the percent yield be?
Solution
i. The ratio of moles of CuSO4 to moles of NaOH is 1:2. This is derived from the balanced chemical equation given.
ii. To determine the limiting reagent, we first need to calculate the number of moles for each reactant.
The molar mass of CuSO4 is approximately 159.61 g/mol and the molar mass of NaOH is approximately 40.00 g/mol.
So, the number of moles of CuSO4 = 638.44 g / 159.61 g/mol = 4.00 mol And the number of moles of NaOH = 240.0 g / 40.00 g/mol = 6.00 mol
Since the ratio of CuSO4 to NaOH is 1:2, we need twice as many moles of NaOH for the reaction to go to completion. However, we have less than this (4.00 mol of CuSO4 would need 8.00 mol of NaOH, but we only have 6.00 mol). Therefore, NaOH is the limiting reagent.
iii. The balanced equation tells us that 1 mole of CuSO4 reacts with 2 moles of NaOH to produce 1 mole of Cu(OH)2. Therefore, the number of moles of Cu(OH)2 produced is equal to the number of moles of the limiting reagent (NaOH), which is 6.00 mol.
The molar mass of Cu(OH)2 is approximately 97.56 g/mol. So, the mass of Cu(OH)2 that should precipitate out is 6.00 mol * 97.56 g/mol = 585.36 g.
iv. The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%.
So, the percent yield = (174.6 g / 585.36 g) * 100% = 29.83%.
Similar Questions
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