Question

# PbCO3(s) + 2HNO3(aq) -> Pb(NO3)2(aq) + H2O(l) + CO2(g) Pb(NO3)2(aq) + 2HCl(aq) -> 2HNO3(aq) + PbCl2(s)...

PbCO3(s) + 2HNO3(aq) -> Pb(NO3)2(aq) + H2O(l) + CO2(g)

Pb(NO3)2(aq) + 2HCl(aq) -> 2HNO3(aq) + PbCl2(s)

(A) If a student starts with 4.000 g of lead(II) carbonate for the first reaction and all other reagents are added in excess, what is the theoretical yield of lead(II) chloride solid?

(B) If the student isolates 3.571 g of lead(II) chloride, what is the percent yield?

A) PbCO3(s) + 2HNO3(aq) -> Pb(NO3)2(aq) + H2O(l) + CO2(g)

As per the stoichiometry of the equations, 1 mole of lead carbonate will give one mole of lead nitrate. And one mole of lead nitrate will give one mole of PbCl2

Now the amount of lead carbonate taken = 4 grams

so moles of lead carbonate = Mass / Molecular weight of lead carbonate

Moles of lead carbonate = 4 / 267 = 0.015 moles

so we will obtain 0.015 moles of lead chloride

Molecular weight of lead chloride = 278

so mass of lead chloride obtained = Moles X molecular weight = 0.015 x 278 = 4.17 grams

So the theroetical yeild will be 4.17 grams

B) the actual yield = 3.571 grams

So % yield = Actual yield X 100 / Theroretical yield = 85.64 %

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