Answer to Question #104431 in General Chemistry for A

Question #104431

Hi,May I please ask I’m doing an experiment on quantitative analysis of calcium carbonate and before doing the experiment I have to answer this: A mass or 0.150g of a sample of commercial CaCo3 was transferred into a 250mL Erlenmeyer flask.50.00 mL of 0.2275M HCL was pipetted into a sample and the solution was boiled for two minutes.The excess acid was titration with 0.1158M NaOH.The average NaOH Titre was 24.19 mL.Calculate the mass of CaCO3 in grams,present in the commercial sample.Please help me?I am unable to submit order.Please understand?Only this time please?

Expert's answer

CaCO3 + 2HCl= CaCl2 + H2O

NaOH+ HCl= NaCl + H2O

0.02419 x 0.1158 = 0.002801202

0.2275 x 0.05 = 0.011375.

0.011375-0.002801202=0.008573798

100 x 0.008573798 /2 = 0.4286899 g of CaCO3.