Question

Using the equation CaCO3 (s) + 2 HCl (aq) arrow CaCl2 (aq) + CO2 (g) + H2O(l)calculate the mass of CaCO3 required to completely react to copletely react with 175 mL of 0.325 M HCl solution.

Answer 1

In this question, we need to find the mass of CaCO3 from the given reaction and the given information about HCl in the reaction, first step will be setting up the properly balanced equation:

CaCO3 + 2 HCl -> CaCl2 + CO2 + H2O

Now we have:

175 mL of HCl

0.325 M of HCl

We need to find the number of moles of HCl using the Molarity formula, which is:

M = n/V

Where:

n is the number of moles

V is the volume in Liters

0.325 = n/0.175

n = 0.057 moles of HCl

Now according to the molar ratio between CaCO3 and HCl, we have 1 mol of CaCO3 for every 2 moles of HCl, what if we have 0.057 moles of HCl:

1 CaCO3 = 2 HCl

x CaCO3 = 0.057 HCl

x = 0.03 moles of CaCO3

Now we use the number of moles found of CaCO3 and its molar mass, 100.1g/mol, to find the final mass of the compound:

100.1g = 1 mol

x grams = 0.03 moles

x = 3.003 grams of CaCO3