1 answer

The following series of reactions were carried out. PbCO_3(s) + 2HNO_3(aq) rightarrow Pb(NO_3)_2(aq) + H_2O(l) +...

Question:

2) The following series of reactions were carried out. PbCo3(s) 2HNO, (aq) Pb(NO3), (aq) H20 (l) Co2(g) Pb(Nor).(aq) 2HCI (aq) 2HNO3(aq) PbCl2(s) starts with 4.000 of lead(ID 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?
The following series of reactions were carried out. PbCO_3(s) + 2HNO_3(aq) rightarrow Pb(NO_3)_2(aq) + H_2O(l) + CO_2(g) Pb(NO_3)_2(aq) + 2HCl(aq) rightarrow 2HNO_3(aq) + PbCl_2(s) 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? If the student isolates 3.571 g of lead(II) chloride, what is the percent yield?

Answers

The reactions are

PbCo3+2HNO3---------> Pb(NO3)2 + H2O(l) +CO2(g)

Pb(NO3)2 + 2HCl ------>PBCl2 +2HNO3

when added, the reactions

Pb(CO3)+ 2HCL-------> PbCl2 + H2O(l) +CO2(g)

Molar mas of Pb(CO3) =267 and that of PbCl2= 278

moles: Pb(CO3)2= 4/267=0.015

from the final reaction, moles of PbCl2 formed from 1 mole of Pb(CO3)2, the limiting reactant are =0.015

Mass of PbCl2 formed = moles* Molar mass =0.015*278= 4.17 gm. This is the theoretical yield

% yield = 100*(actual yield/theoretical yield)=100*3.571/4.17=85.64%

.

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