Answer to Question #296167 in General Chemistry for acks

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)}

Answer 1

From the equation:

3 mol Ba(NO3)2 react with 1 mol Al2(SO4)3

Mol Al2(SO4)3 25.0 mL of 0.350 M solution

Mol = 25.0 mL / 1000 mL/L ,* 0.350 mol /L = 0.00875 mol

This will require 0.00875*3 = 0.02625 mol Ba(NO3)2

The Ba(NO3)2 solution is 0.280 M

1000 mL contains 0.280 mol

Volume that contains 0.02625 mol = 0.02625 mol / 0.280 mol * 1000 mL = 93.75 mL