In one experiment, a student added a total of 15.00 mL of 0.5471 M HCl to 1.4741

Question

In one experiment, a student added a total of 15.00 mL of 0.5471 M HCl to 1.4741 g to of an unknown which contained CaCO3 (molar mass = 100.09 g/mol). The resulting solution required 12.36 mL of 0.1211 M NaOH to back titrate the excess HCl. What was the percent CaCO3 in the unknown? Report your answer to two (2) decimal places. Do not include units (%) in your answer.

Explanation / Answer

Molarity = moles/ volume(liteer) moles = molarity *& volume Moles of HCl= 0.547*(15/1000) = 0.008205 mole HCl Moles of NaOH = 0.1211 M * (12.36/1000) =0.00150 mol NaOH So moles of excess HCl was 0.00150. So moles of HCl that copmbines with CaCO3 was (0.008205-0.00150)=0.006705 mol CaCO3 + 2 HCl =====> CaCL2 + CO2 + H2O so 0.006705 mol HCl * (1 mol CaCO3/ 2 mol Hcl)*(100.09g CaCO3/ 1mol CaCo3) 0.3336 g of CaCO3 So in 1.4741g of the sample there were 0..336 g CaCO3 So mass % = (0.336/1.4714)*100 = 22.835 = 22.84% CaCO3

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