iv. If only 174.6 g of Cu(OH)2 precipitate were actually collected from the reaction, what would the percent yield be? SHOW ALL WORK!

The percent yield is   59.4%

Firstly write the balanced chemical equation:

CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(aq) + Na₂SO₄ (aq)

Ratio of moles for CuSO₄ to NaOH is 1:2

To determine the limiting reagent:

For CuSO₄:

1 mole of CuSO₄ produces 1 mole of Cu(OH)₂

4 moles of CuSO₄ will produce 4 moles of Cu(OH)₂

For NaOH:

2 moles of NaOH yields 1 mole of Cu(OH)₂

6 moles of NaOH yields moles of Cu(OH)₂

Hence, NaOH is the limiting reactant.

Therefore, mass of Cu(OH)₂ =

% yield = actual yield/ theoretical yield × 100

% yield=

% yield = 59.4%

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59.4%

Explanation:

CuSO4(aq) + 2NaOH(aq) > Cu(OH)2(aq) + Na2SO4 (aq)

Ratio of molesCuSO4 to NaOH is 1:2

To determine the limiting reagent;

For CuSO4

Number of moles= 638.44g/160 = 4 moles

1 mole of CuSO4 yields 1 mole of Cu(OH)2

4 moles of CuSO4 also yields 4 moles of Cu(OH)2

For NaOH

Number of moles= 240g/40g/mol= 6 moles

2 moles of NaOH yields 1 mole of Cu(OH)2

6 moles of NaOH yields 6× 1/2 = 3 moles of Cu(OH)2

Hence NaOH is the limiting reactant.

Hence mass of Cu(OH)2 = 3 × 98 g/mol

= 294 g

% yield = actual yield/ theoretical yield × 100

% yield= 174.6 /294 × 100/1

% yield = 59.4%

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante