Answer to Question #296696 in General Chemistry for yui

Question

SOLUTION STOICHIOMETRY: Using the given BALANCED EQUATION, and calculate for what is asked in each number. Show complete solutions and write appropriate units for your final answers.

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 BaSO4 (s) + 2 Al(NO3)3 (aq

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 HNO3 (aq) + Ba(OH)2 (aq) --> 2 H2O (aq) + Ba(NO3)2 (aq)

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

Answer 1

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

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

3 Ba(NO3)2 (aq) + Al2(SO4)3 (aq) --> 3 BaSO4 (s) + 2 Al(NO3)3 (aq