Please help me with this chemistry calculation?

Question

Composition of a Mixture by Acid-Base Titration

A 1.00-g solid sample containing a mixture of table salt (NaCl) and citric acid (H3C6H5O7, a triprotic acid) is dissolved in 15 mL of water.
Titration of the acid solution requires 14.90 mL of 0.3196 M NaOH solution to reach the endpoint.
Calculate the mass percent H3C6H5O7 in the solid mixture.



I don't care when you answer... even till next week... but please just help.. I don't get it =(

Answer 1

14.90 mL of 0.3196 M NaOH solution contains 0.004762 Mole of NaOH. Since 3 moles of NaOH is needed to titrate 1 mole of citric acid to the end point, 0.004762 Mole of NaOH can titrate 0.001587 Mole of citric acid, which is 0.3048 gram.
Therefore the mass percent H3C6H5O7 in the solid mixture is 30.5%.

Answer 2

Ferric Oxide = Fe2O3 reason backwards... a million.00g Fe2O3 would desire to incorporate some mass of Fe, we would desire to confirm that. We accomplish that by way of first calculating the p.c. Mass of Iron in Fe2O3. Fe = fifty 5.80 5 g/mol O = sixteen.00 g/mol 2 * (fifty 5.80 5) + 3 * (sixteen.00) = 159.7 g/mol %Fe = 2 * (fifty 5.80 5 g/mol) / 159.7 g/mol = 0.6994 this implies the a million.00g of Precipitate, incorporates 0.6994 g of Fe. Now, the unique cord itself became 0.800g as a thank you to get the %Fe interior the cord, we divide the mass of Iron (obtained from the precipitate) by way of the mass of the cord: 0.6994 g / 0.800 g = .874 Multiply this answer by way of one hundred to make it a share of 87.4%