Answer:
The percent yield of lead(II) chloride is 88,8%
Explanation:
2HCl + Pb(NO3)2 → 2HNO3 + PbCl2
By stoichiometry, relation between Pb(NO3)2 and PbCl2 IS 1:1 mol
So 1 mol of Pb(NO3)2, makes 1 mol of PbCl2
I use 870 g of lead (II) nitrate, which means 2,63 moles
Molar mass - Pb(NO3)2-: 331,2 g/mol
Mass/Molar mass : Moles → 810 g /331 g/m = 2,63 moles
In conclussion, I obtanied 2,63 moles of PbCl2 but this result is at 100% yield of reaction, as I only formed 650 gr of PbCl2 I have to find out my yield.
Molar mass -PbCl2-: 278,1 g/mol
Mass/Molar mass : Moles → 650g / 278,1 g/mol = 2,34 moles
2,63 moles ______ 100 %
2,34 moles ______ (2,34 . 100) /2,63 = 88,8 %
