Question

b. 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?

Answer 1

Answer:

i. 1 : 2

ii. NaOH is the limiting reagent

iii. 292.5g

iv. 59.69%

Explanation:

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Answer 2

1. The ratio of CuSO₄ to NaOH is 1 : 2

2. The limiting reactant is NaOH.

3. The mass of Cu(OH)₂ that will precipitate out is 292.5 g.

4. The percentage yield of Cu(OH)₂ is 59.7%

1. Determination of the ratio of CuSO₄ to NaOH.

CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄

From the balanced equation above, we can see that the ratio of CuSO₄ to NaOH is 1 : 2

2. Determination of the limiting reactant.

CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄

Molar mass of CuSO₄ = 63.5 + 32 + (16×4) = 159.5 g/mol

Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Mass of NaOH from the balanced equation = 2 × 40 = 80 g

From the balanced equation above,

159.5 g of CuSO₄ reacted with 80 g of NaOH.

Therefore,

638.44 g of CuSO₄ will react with = (638.44 × 80) / 159.5 = 320.22 g of NaOH.

From the calculations made above, we can see that a higher mass (i.e 320.22 g) of NaOH than what was given (i.e 240 g) is needed to react completely with 638.44 g of CuSO₄.

Therefore, NaOH is the limiting reactant.

3. Determination of the mass of Cu(OH)₂ that will precipitate out.

CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄

Molar mass of Cu(OH)₂ = 63.5 + 2(16 + 1) = 97.5 g/mol

Mass of Cu(OH)₂ from the balanced equation = 1 × 97.5 = 97.5 g

From the balanced equation above,

80 g of NaOH reacted to produce 97.5 g of Cu(OH)₂.

Therefore,

240 g of NaOH will react to produce = (240 × 97.5) / 80 = 292.5 g of Cu(OH)₂

Thus, 292.5 g of Cu(OH)₂ precipitated out of the reaction.

4. Determination of the percentage yield.

• Actual yield of Cu(OH)₂ = 174.6 g

• Theoretical yield of Cu(OH)₂ = 292.5 g

• Percentage yield of Cu(OH)₂ =?

Percentage yield = (Actual /Theoretical) × 100

Percentage yield of Cu(OH)₂ = (174.6 / 292.5) × 100

Percentage yield of Cu(OH)₂ = 59.7%

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