Answer to Question #249575 in General Chemistry for mas

Question

How many mL of 0.548 M HCl are needed to dissolve 5.89 g of CaCO₃?

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

Answer 1

Molar mass of CaCO3= 100.0869 g/mol

5.89/100.0869= 0.059mol

Mole ratio HCl: CaCO3= 2:1

0.059×2= 0.108mol

Molar mass of HCl = 18.015

0.108×18.015= 1.95g

0.548×18.015=9.87g

9.87/1.95

= 5.063g