a sample of magnesium carbonate weighing 1 gram was made with the volume of 50 milliliters of nitric acid 0.2 n and needed to equalize the excess of acid 35 milliliters of sodium hydroxide 0.1 n to calculate the percentage purity of the sample?

Answer 1

I will assume that n stands for normality but since these are 1:1 acid and 1:1 base, that is the same as molarity = M. Then mL x M = millimoles.
2HNO3 + MgCO3 ==> Mg(NO3)2 + CO2 + H2O
millimoles HNO3 initially = mL x M = 50 x 0.2 = 10
millimoles NaOH added after MgCO3 reaction = 35 x 0.1 = 3.5
millimoles HNO3 in excess = 10 - 3.5 = 6.5
millimoles MgCO3 that reacted = 1/2*6.5 = 3.25 millimoles = 0.00325 moles = ? grams. g = mols x molar mass = 0.00325 x (84.3) = 0.274 g
% purity = (0.274 g/1 g) *100 = ?