Answer to Question #125163 in General Chemistry for mike

Question

A 0.9030 gram sample of M(OH)2(aq) was mixed with 20.00mL of 2.000M HCl and enough water added to make 100.00mL of solution.
M(OH)2(aq) + 2HCl(aq) ⇨ MCl2(aq) + 2H2O(l)

Answer 1

Moles of HCl added = 0.02000 L x 2.000 M = 0.04000 mole HCl

Moles of NaOH added to the 10.00 mL aliquot = 0.01764 L x 0.05121 M = 0.000903168

Equivalent number of moles of NaOH required for the original 100.00 mL solution = 0.000903168 x 10 = 0.00903168

Moles of HCl that reacted with M(OH)2 in the 100.00 mL solution = 0.04000 - 0.00903 = 0.03097

Moles of M(OH)2 neutralized by 0.03097 moles HCl = 0.03097 / 2 = 0.015485 

Grams of M(OH)2 in the 100.00 mL solution = 0.9030 g

Molar mass of M(OH)2 = 0.9030 g / 0.015485 mole = 58.31 g/mol

58.31 g/mol - molar mass of 2 x OH = 58.31 - (2 x 17.008) = 24.30

Atomic mass of M = 58.31 - 34.02 = 24.3

M = Mg

Mg(OH)2