How many mL of 0.280 M barium nitrate are required to precipitate as barium sulfate all the fulfate ions from 25.0mL of 0.350 M aluminum sulfate? Balanced equation: 3Ba(NO3) 2(aq) + Al2(SO4) 3(aq) + 3BaSO4(s) + 2Al(NO3) 3(aq)

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

Answer sholuld have 3 significant digit : Volume = 93.8 mL required.

1. 3Ba(NO3)2(aq) + Al2(SO4)3(aq) = 3BaSO4(s) + 2Al(NO3)3(aq)
2. 3:1:3:2 ratio
3. 0.025L x 0.350mole/L = 0.00875mole of Al2(SO4)3(aq)
4. 0.280mole/L x aL = 0.00875mole x 3
5. Solve for aL = (0.00875 x 3) / 0.280 = 0.0938L = 93.8ml of 0.280M Ba(NO3)2