How many grams of lead(II) chloride is produced if 13.87 g lead(II) nitrate combines with excess hydrochloric acid to produce lead(II) chloride and nitric acid?
A) 5.82 g
B) 14.33 g
C) 0.086 g
D) 11.64 g
E) 16.52 g

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction is:

Pb(NO3)2 + 2HCl → PbCl2 + 2HNO3

From the equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of HCl to produce 1 mole of PbCl2. Therefore, we need to first convert the given mass of Pb(NO3)2 to moles, and then use the mole ratio to calculate the mass of PbCl2 produced.

1. Calculate moles of Pb(NO3)2:

molar mass of Pb(NO3)2 = 207.2 + 2(14.01 + 3(16.00)) = 331.2 g/mol
moles of Pb(NO3)2 = 13.87 g / 331.2 g/mol = 0.0418 mol

2. Use mole ratio to calculate moles of PbCl2:

moles of PbCl2 = moles of Pb(NO3)2 = 0.0418 mol

3. Calculate mass of PbCl2:

molar mass of PbCl2 = 207.2 + 2(35.45) = 278.1 g/mol
mass of PbCl2 = moles of PbCl2 x molar mass of PbCl2
= 0.0418 mol x 278.1 g/mol = 11.64 g

Therefore, the answer is D) 11.64 g.

How Many Grams of Lead(II) Chloride Is Produced If 13