Explain naoh to back titrate the excess hcl

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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

Reference no: EM13245491

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