what is the mass and the identity of the precipitate that forms when 55.0 ml of 0.100 M BaCl2 reacts with 40.0 ml of 0.150 M Na2CO3?

Answer 1

BaCl2 + Na2CO3 --> BaCO3 + 2NaCl

Barium carbonate is the precipitate.

55ml x 0.1 M = 5.5 mmol Ba

40ml x 0.15 M = 6.0 mmol CO3 Thus limiting reagent is Ba.

Molar mass barium carbonate =

137 + 12 + 3 x 16 = 197 g/mole = 197 mg/mmol

5.5 mmol x 197 mg/mmol = 1083.5 mg

Round for sig figs .... 1080 mg barium carbonate