Calcium carbonate reacts with aqueous HCl to give CaCl_(2) and CO_(2) according to the reaction, CaCO_(3) (s) + 2 HCl (aq) to CaCl_(2) (aq) + CO_(2)(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Calcium carbonate reacts with aqueous HCl to give CaCl_(2) and CO_(2)

1.	Calcium carbonate reacts with aqueous HCl to give CaCl_(2) and CO_(2) according to the reaction, CaCO_(3) (s) + 2 HCl (aq) to CaCl_(2) (aq) + CO_(2)(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
Answer» Solution :Molarity (M) of a solution is given by: `w=(MM.V)/1000` In the present case, M = 0.75, M. = 1 + 35.5 = 36.5, V = 25 ML, w =? Substituting the values, we have `w=(0.75 xx 36.5 xx 25)/1000 = 0.6844 G` Therefore, the mass of HCI present in the given solution is 0.6844 g. The given EQUATION is: `underset("1 mole 100 g")(CaCO_(3)(s)) + underset("2 moles " 2 xx 36.5 = 73 g)(2HCl(aq)) to CaCl_(2)(aq) + CO_(2)(g) + H_(2)(l)` `therefore` 73 g of HCI require for complete REACTION the mass of `CaCO_3 = 100 g` `therefore` 0.6844 g of HCI will require `CaCO_(3) = 100/73 xx 0.6844 = 0.939 g`