Question

What is the mass of the precipitate (BaCO3) that forms when 65 mL of 0.3 M BaCl2 reacts with 60 mL of 0.35 M NaCO3?

Answer 1

moles of BaCl2 = 65 x 0.3 / 1000 = 0.0195

moles of Na2CO3 = 60 x 0.35 / 1000 = 0.021

BaCl2 + Na2CO3   --------------> BaCO3 + 2 NaCl

1                1                                    1

0.0195      0.021

here limiting reagent is BaCl2. because mol ratio of BaCl2 is less.

1 mol BaCl2 ------------>. 1 mol BaCO3

0.0195 mol   ----------------> ?/

moles of BaCO3 = 0.0195

mass of BaCO3 = 0.0195 x 197.3

mass of BaCO3 = 3.85 g BaCO3