Answer to Question #249606 in General Chemistry for maruam

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)

 _____________mL

Answer 1

Molar mass of CaCO3= 100.0869 g/mol

5.89/100.0869= 0.0588mol

Molar mass of HCl= 36.458 g/mol

0.0588×36.458×2= 4.29ml