Answer to Question #183485 in General Chemistry for Krissy

Question

Solution Stoichiometry Problem Set

How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added 

      to 100. mL of 0.400 M potassium chromate? (7 marks)

2.   How many mL of 0.280 M barium nitrate are required to precipitate as barium sulfate all the sulfate  

      ions from 25.0 mL of 0.350 M aluminum sulfate? (4 marks)

Answer 1

Equation of the reaction 2Ag+CrO4=Ag2CrO4

Silver chromate will deposit from solution as a fine brick-red precipitate with alacrity; the material is exceptionally insoluble.

And thus we calculate the equivalent quantities of silver ion

 and chromate ion

M=V×Molarity

Moles of silver nitrate = 0.15×0.5 = 0.75moles

Moles of Potassium Chromate = 0.1×0.4 = 0.4moles

Mass = moles × molar mass

Mass=0.5×0.75×331.73= 124.4g

Ans = 124.4g

2. 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

Ans = 93.75mL