What mass of an 85 % pure sample of Mg(OH)₂ is required to completely neutralize 250 mL of a 0.4 M HNO₃ solution according to the given reaction? Mg(OH)_ 2(s) + 2HNO_ 3(aq) Mg(NO₃)_ 2(aq) + 2H₂O_ (l) [Calculate up to the second place of decimal point. Atomic masses: Mg = 24 , O = 16 , H = 1 ]

Step-by-step solution

Moles of HNO₃ = M V = 0.4 250 1000 = 0.1 mol From the balanced chemical equation, 1 mol of Mg(OH)₂ neutralizes 2 mol of HNO₃ . Moles of Mg(OH)₂ required = 0.1 2 = 0.05 mol Molar mass of Mg(OH)₂ = 24 + 2 (16 + 1) = 58 g/mol Mass of pure Mg(OH)₂ required = 0.05 58 = 2.9 g Since the sample is 85 % pure, the mass of the sample required is given by: m 85 100 = 2.9 m = 2.9 100 85 = 3.41 g Answer: 3.41 g