Question

a mixture of bases can sometimes be the active ingredient in antacid tablets. If .4896 g of a mixture of Al(OH)3 and Mg(OH)2 is neutralized with 17.27 mL of 1.000 M HNO3, what is the mass % of Al(OH)3 in the mixture?

Answer 1

m = 04896 g of Al(OH)3 & Mg(OH)2

V = 17.27 ml of 1 M HNO3

find Al(OH)3

The neutralization reactions

Mg(OH)2 + 2HNO3 = Mg(NO3)2 + 2H2O

Al(OH)3+ 3HNO3 = Al(NO3)3 + 3H2O

ratio is 2:3; in mole

so

mol of acid used = M*V = 17.27/1000*1 = 0.01727 mol of acid used

2:1 ratio for yMg --> 1/2*0.01727 = 0.008635 mol of Mg

3:1 are Al --> 1/3*0.01727 = 0.00575 mol of Al

therefore we have 1:! ratio between Al(OH)3 and Al

then

0.00575 mol of Al

then

% mas sof Al(OH)3 = mass of Al(OH)3 / total mass * 100

MW Al(OH)3 = 78.0036

mass of Al(OH)3 = 0.00575* 78.0036 = 0.4485207 g

then

% mass = 0.4485207/0.4896 *100 = 91.6096%