Asked by dante
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?
Answers
Answered by
DrBob222
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 = ?
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 = ?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.