What volume of 0.200 M HCl will react with 0.2500 g of CaCO3? CaCO3 (s) + 2 HCl (aq) ➔ CaCl2(aq) + H2O (l) + CO3 (g).

0.02498 L

Explanation:

CaCO₃  +  2 HCl  ➔  CaCl₂  +  H₂O  +  CO₂

You have to use stoichiometry.  According to the chemical equation, for every one mole of CaCO₃, two moles of HCl is needed for the reaction to occur.  Before you can use this relation you need to convert grams of CaCO₃ to moles.  To convert, you need to use the molar mass.

Molar mass of CaCO₃ = 100.086 g/mol

(0.2500 g)/(100.086 g/mol) = 0.002498 mol CaCO₃

Now using the relation of 1 mol of CaCO₃ for every 2 mol of HCl, convert moles of CaCO₃ to moles of HCl.

(0.002498 mol CaCO₃)  (2 mol HCl)/(1 mol CaCO₃) = 0.004996 mol HCl

Since the molarity of the solution of HCl is 0.200 M (mol/L), you have to divide the amount of moles needed by the molarity of the solution.

(0.004996 mol HCl)/(0.200 M) = 0.02498 L

You will need 0.02498 L to react with 0.2500 g of CaCO₃.

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