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Answer:
261.337 g/mol
Explanation:
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Pb(NO₃)₂ ⇒limiting reactant
moles PbI₂ = 1.36 x 10⁻³
% yield = 87.72%
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)
Required
- moles of PbI₂
- Limiting reactant
- % yield
Solution
Balanced equation :
Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
mol Pb(NO₃)₂ :
= 0.45 : 331 g/mol
= 1.36 x 10⁻³
mol NaI :
= 250 ml x 0.25 M
= 0.0625
Limiting reactant (mol : coefficient)
Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³
NaI : 0.0625 : 2 = 0.03125
Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)
moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)
Mass of PbI₂ :
= mol x MW
= 1.36 x 10⁻³ x 461,01 g/mol
= 0.627 g
% yield = 0.55/0.627 x 100% = 87.72%