45g of Calcium carbonate is added to a 1.5L solution of hydrochloric acid with a

Question

45g of Calcium carbonate is added to a 1.5L solution of hydrochloric acid with a density of 1.13 gHCl/mL. Assuming volume remains constant, what will the molar concentration of HCl be after the calcium carbonate is added? 45g of Calcium carbonate is added to a 1.5L solution of hydrochloric acid with a density of 1.13 gHCl/mL. Assuming volume remains constant, what will the molar concentration of HCl be after the calcium carbonate is added? 45g of Calcium carbonate is added to a 1.5L solution of hydrochloric acid with a density of 1.13 gHCl/mL. Assuming volume remains constant, what will the molar concentration of HCl be after the calcium carbonate is added?

Explanation / Answer

m = 45 g of CaCO3

V = 1.5 L of 1.13 g HCL/mL

mol of CaCO3 = mass/MW = 45/100 = 0.45 mol

CaCO3 + 2HCl = H2O + CO2 + CaCl2

mol of CaCO3 --> 2 mol of HCl

0.45 mol of CaCO3 = 0.90 mol of HCl

[HCl] = mol/V= 0.90 / 1.5 = 0.6 M

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