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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%
You need to use the equation PV = nRT
The only “trick” in these problems is you need to use the correct UNITS. Especially for the “R” value.
If you get the units (eg temps in Celsius or Kelvin) corresponding to the correct “matching” units of the other variables.
You need to convert the temp into Kelvin first