yield = 80 %
Explanation:
We have the following chemical reaction:
Cu(NO₃)₂ (aq) + Na₂S (aq) → 2 NaNO₃ (aq) + CuS (s)
where:
(aq) - aqueous
(s) - solid
number of moles = mass / molecular weight
number of moles of Cu(NO₃)₂ = 469 / 187.5 = 2.5 moles
number of moles of Na₂S = 156 / 78 = 2 moles
number of moles of NaNO₃ = 272/ 85 = 3.2 moles (practical)
We see that from the chemical reaction 1 mole of Cu(NO₃)₂ will react with 1 mole of Na₂S. Because we have 2.5 moles of Cu(NO₃)₂ and 2 moles of Na₂S, the limiting reactant will be Na₂S.
Taking this into account, we devise the following reasoning:
if 1 mole of Na₂S produces 2 moles of NaNO₃
then 2 moles of Na₂S produces X moles of NaNO₃
X = (2 × 2) / 1 = 4 moles of NaNO₃ (theoretical)
The yield of the reaction will be:
yield = (practical quantity / theoretical quantity) × 100
yield = (3.2 / 4) × 100 = 80 %
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yield of a chemical reaction
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