Question

A mixture of bases can sometimes be the active ingredients in antacid tablets. If 0.4864g of a mixture of Al(OH)3 and Mg(OH)2 is neutralized with 17.32 mL of 1.00 M HNO3, what is the mass % of Al(OH)3 in the mixture?

Answer 1

Al(OH)₃ + 3HNO₃ ---> Al(NO₃)₃ + 3H₂O

Mg(OH)₂ + 2HNO₃ ------> Mg(NO₃)₂ + 2H₂O

Let there be 'x' g of Al(OH)₃

Thus, mass of Mg(OH)₂ = (0.4864-x) g

Molar mass of Al(OH)₃ = 78 g/mole

Molar mass of Mg(OH)₂= 58 g/mole

Thus, moles of Al(OH)₃ reacting = x/78

moles of Mg(OH)₂ reacting = (0.4864-x)/58

Moles of HNO₃ reacting = molarity*volume of solution in litres = 1*0.01732 = 0.01732

Thus, 3*moles of Al(OH)₃ + 2*moles of Mg(OH)₂ = moles of HNO₃

or, (3x/78) + (0.9728-2x/58) = 0.01732

or, (18x + 75.8784)/4524 = 0.01732

or, x = 0.138 g

Thus, mass % of Al(OH)₃ = (0.138/0.4864)*100 = 28.295%