Answer to Question #295980 in General Chemistry for johny

Question

1.How many mL of 0.250 M barium nitrate are required to precipitate as barium sulfate all the sulfate ions from 50 mL of 0.350 M aluminum sulfate?

3 Ba(NO₃)_{2 (aq)}           +       Al₂(SO₄)_3 (aq)     -->   3 BaSO_{4 (s)}  + 2 Al(NO₃)_{3 (aq)}

1. 2.5 mL of nitric acid, HNO₃, are required to react with 50 mL of a 1.25 M Barium hydroxide, Ba(OH)₂ solution:

2 HNO_3 (aq)           +      Ba(OH)_{2 (aq)} -->    2 H₂O _(aq)    +      Ba(NO₃)_{2 (aq)}

Determine the concentration (Molarity) of the nitric acid solution (HNO₃).

Answer 1

Molar mass of barium nitrate 261.337

0.25×261.337=65.33g

Molar mass of aluminum sulfate = 342.15 g/mol

0.35×342.15=119.8

= 119.8/65.33=1.83moles