Question

Part 1. A 13 ml sample of a 1.38 M potassium chloride solution is mixed with 52.9 ml of a 1.43 M lead(II) nitrate solution and a precipitate forms. The solid is collected, dried, and found to have a mass of 0.9 g. What is the percent yield? Enter to 2 decimal places.

Part 2. What is the net ionic equation of barium hydroxide + hydrofluoric acid?

Answer 1

1)

find number of mol of KCl:

volume , V = 13 mL

= 1.3*10^-2 L

number of mol,

n = Molarity * Volume

= 1.38*0.013

= 1.794*10^-2 mol

This is number of mol of KCl

find number of mol of Pb(NO3)2:

volume , V = 52.9 mL

= 5.29*10^-2 L

number of mol,

n = Molarity * Volume

= 1.43*0.0529

= 7.565*10^-2 mol

This is number of mol of Pb(NO3)2

Balanced chemical equation is:

2 KCl + Pb(NO3)2 ---> PbCl2 + 2 KNO3

2 mol of KCl reacts with 1 mol of Pb(NO3)2

for 0.01794 mol of KCl, 0.00897 mol of Pb(NO3)2 is required

But we have 0.075647 mol of Pb(NO3)2

so, KCl is limiting reagent

we will use KCl in further calculation

Molar mass of PbCl2,

MM = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

According to balanced equation

mol of PbCl2 formed = (1/2)* moles of KCl

= (1/2)*0.01794

= 0.00897 mol

mass of PbCl2 = number of mol * molar mass

= 8.97*10^-3*2.781*10^2

= 2.495 g

% yield = actual mass*100/theoretical mass

= 0.9*100/2.495

= 36.1%

2)

Ba(OH)2 (aq) + HF (aq) —> BaF2 (aq) + H2O (l)

The net ionic equation will be:

H+ (aq) + OH- (aq) —> H2O (l)