Calculate the amount of HCl in grams required to react with 3.75 g of CaCO3 according to the following reaction: CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

The correct answer is 2.75 grams of HCl.

Explanation:

The given balanced equation is:  

CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams

The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,  

= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.