yield = 52.23 %
Explanation:
We have the following chemical reaction:
2 Al (s) + 2 KOH (aq) + 4 H₂SO₄ (aq) + 10 H₂O → 2 KAl(SO₄)₂·12 (H₂O) (s) + 3 H₂ (g)
mass of aluminium = mass of bottle with aluminium pieces - bottle mass
mass of aluminium = 10.8955 - 9.8981 = 0.9974 g
mass of alum = mass of bottle with final product - bottle mass
mass of alum = 19.0414 - 9.8981 = 9.1433 g
number of moles = mass / molecular weight
number of moles of aluminium = 0.9974 / 27 = 0.03694 moles
number of moles of alum (practical) = 9.1433 / 474 = 0.01929 moles
To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:
if 2 moles of aluminium produce 2 moles of alum
then 0.03694 moles of aluminium produce X moles of alum
X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)
yield = (practical quantity / theoretical quantity) × 100
yield = (0.01929 / 0.03694) × 100
yield = 52.23 %
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