Equation: Pb(NO3)2(aq) + K2(CO3) --> PbCO3(s) + 2KNO3(aq) Net Ionic Equation: Pb2+ + CO32- --> PbCO3(s)...

Question

Question

Equation: Pb(NO₃)₂(aq) + K₂(CO₃) --> PbCO₃(s) + 2KNO₃(aq)

Net Ionic Equation: Pb²⁺ + CO₃^2- --> PbCO₃(s)

lead (ii) nitrate concentration = 0.15 M, potassium carbonate concentration = 0.20 M

Determine the percent yield if 0.1150 L of each reactant were allowed to react, and a mass of 5.0012 g of solid were obtained.

Answer 1

Answer #1

Limiting reagent is lead nitrate...

Lead Nitrate of 0.1150 L = 0.20*0.1150 mol =0.023 mol

Thus , Lead carbonate should produce at 0.023 mol in 100% theoretical yield...

0.01725 mol of Lead carbonate = 0.023*267.21 g Lead Carbonate =6.145 g lead carbonate .

(267.21 g is the molecular weight of Lead carbonate) .

Thus - Yield -

5.0012 Y = X 100 6.145 %

Therefore, Y=81.38%

Answer 2

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