Chemistry Stoichiometry

Question

(3.71x10^2) mL of 0.175-mol/L Pb(NO₃)₂ (aq) is mixed with excess KBr (aq). Calculate the mass of the precipitate.

Note: Your answer is assumed to be reduced to the highest power possible

Answer 1

We start our problem with a balanced reaction equation as follows:

Pb(NO₃)₂ + 2 KBr → PbBr₂ + 2 KNO₃

The precipitate we'll see is the lead bromide as its solubility in water (K_sp) at STP is 1.86 x10^-5.

We are given an initial amount of lead nitrate: 371 mL of 0.175 M. We calculate the number of moles of lead nitrate to be:

0.175 x 0.371 = 0.0649 moles lead nitrate. They didn't give us a finite volume, so I will assume 1L total volume.

After mixing, our lead bromide reaction looks like:

PbBr₂ → Pb²⁺ + 2 Br^-

We know the K_sp of lead bromide is

K_sp = [Pb²⁺][Br^-]²/[PbBr₂] = 1.86x10^-5

Since we know how much PbBr₂ was transformed, we can treat the K_sp like an equation, where:

1.86x10^-5 = x³/0.0649

x = [0.0106]

This means that of the initial 0.0649 moles of lead nitrate, only 0.0106 moles stayed in solution as lead bromide. To calculate how much precipitated:

0.0649 - 0.0106 = 0.0523 moles lead bromide precipitated. To calculate grams from moles:

0.0523 moles * 367.01 g/mole = 19.91 g PbBr₂ precipitates

EDIT: from a comment, if we do not assume ANY solubility rules or K_sp and use only stoichiometry rules, we convert the given volume of lead nitrate to grams to find the total precipitate:

0.371 L * 0.175 M = 0.065 moles lead nitrate, which is an equivalent number of moles of lead bromide. Therefore there are:

0.065 moles * 367.01 g/mole lead bromide = 23.83 g PbBr₂ precipitates