Question

A 25 ml sample of 1.2 molar potassium chloride mix with 15 ml of 0.90 molar barium nitrate solution and precipitate reaction occurs twice case LX + BA no3s aqueous bacl2 solid + 2ks what is the practical yield percentage yield mass is 2.45 g

Answer 1

Percentage yield = (actual yield /
theoretical yield) × 100%
Percentage yield = (2.45 g / 2.81 g) x
100%
Percentage yield = 87.2%
Therefore, the practical yield percentage yield is 87.2%, and the mass of the BaCI2 produced is 2.81 g.


Moles of KCI = volume (in L) X
concentration
Moles of KCI = (25/1000) L x 1.2 mol/L
= 0.03 mol
Moles of Ba(NO32 = volume (in L) X
concentration
Moles of Ba(NO3)2 = (15/1000) L x
0.90 mol/L = 0.0135 mol

Moles of BaCI2 formed = 0.0135 mol
The molar mass of BaCI2 is 208.23 g/ mol, so the mass of BaCI2 produced is:
Mass of BaCI2 = moles of BaC12 x
molar mass of BaCI2
Mass of BaCI2 = 0.0135 mol x 208.23
g/mol
Mass of BaCI2 = 2.81 g