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Sagot :
Answer:
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%
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:
[tex]\text{Number of moles}= \frac{ 638.44g}{160}= 4 \text{moles}[/tex]
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:
[tex]\text{Number of moles}= \frac{240g}{40g/mol} = 6 \text{moles}[/tex]
2 moles of NaOH yields 1 mole of Cu(OH)₂
6 moles of NaOH yields [tex]6*\frac{1}{2} = 3[/tex] moles of Cu(OH)₂
Hence, NaOH is the limiting reactant.
Therefore, mass of Cu(OH)₂ = [tex]3 * 98 g/mol= 294 g[/tex]
% yield = actual yield/ theoretical yield × 100
% yield= [tex]\frac{174.6}{294}* 100[/tex]
% yield = 59.4%
Learn more:
brainly.com/question/2442404
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