Question

What is the mass of the precipitate (BaCO3) that forms when 65 mL of 0.3 M...

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?

Homework Answers

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

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