Question

An industrial vat contains 650 grams of solid lead(II) chloride formed from a reaction of 870 grams of lead(II) nitrate with excess hydrochloric acid. This is the equation 2HCl + Pb(NO3)2 = 2HNO3 + PbCl2 what is the percent yield of lead(II) chloride?

Answer 1

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 %

Answer 2

Answer:

89 for PLATO users

Explanation: