need help with last section !

Identify the correct mole to mole ratio for each type of acid when reacting with a strong base such as NaOH monoprotic acid (HX) 1 mol acid: 1 mor base diprotic acid (H2A) 1 mol acid 2 mol base triprotic acid (H3A) 1 mol acid 3 mol base Given the following generic equations: a. g of acid =1000inkofNaOH molar mass of NaOH2isolenofNOOH1moteofaral molar mass of acid b. g of acid =1000melNaOH molarity of NaOH2mileofNaOiI1maleofari molar mass of acid c. g of acid =1090mLofNiOH molarity of NaOH2motisonNnOH1mateofari4 d. g of acid =mL of NaOH molarity of NaOH2misesofNoOiI1meteofarid molar mass of acid e. g of acid =1000ml.ofNaOHI molarity of NaOH1moleofNaOif2moleofarif molar mass of acid Which generic equation should you use to determine the mass of a diprotic acid analyte, in grams, when the diprotic acid is titrated with NaOH? 33.44mL of 0.103MNaOH was needed to titrate a solid tartaric acid sample. Tartaric acid is a dinmtin acid with molar mass of 150.09q/mol. Given the following equations: Which generic equation should you use to determine the mass of a diprotic acid analyte, in grams, when the diprotic acid is titrated with NaOH ? 33.44 mL of 0.103MNaOH was needed to titrate a solid tartaric acid sample. Tartaric acid is a diprotic acid with molar mass of 150.09g/mol. Given the following equations: a. 33.44mL1000mL1L1L0.103mol1moleNaOH2moleoftartaricacid1mol160.09g=g of acid b. 33.44mL1L0.103mol2moleNaOH1moletartaticadd150.09g/mol=g of acid c. 33.44mL1000mL1L1mol40.00g2moleNaOH1moleoftartaricacid150.09g/mol=g of acid d. g of acid =33.44mL1000mL1L1L0.103mol2moleNaOH1moletartaricacli=g of acid e. g of acid =33.44mL1000mL1L1L0.103mol2moleNaOH1moleoftartaricacid1mol150.00g=g of acid Which equation would you use to calculate how many grams of tartaric acid was in the sample? (2pts) Carrying out the Calculations (1pts) How many grams of tartaric acid were present in the sample? (1pts) How many grams of tartaric acid were present in the sample? The initial mass of the tartaric acid sample was measured to be 0.333g, therefore it was not a pure gample. (3pts) Based on the mass of tartaric acid determined by titration and the mass of the original sample, what percentage of the sample was actually tartaric acid