Question

What volume in ml of 1.5 M HCl is needed to completely react with 92 grams of the CaCO₃?

Answer 1

first calculate the moles of calcium carbonate

Moles = $\frac{Mass}{molar mass} =$ $\frac{92}{100.0869}=0.917$ moles

the equation of reaction gives a mole ratio between calcium carbonate and hydrochloric acid as 1:1

CaCO₃ + HCL = CaCl₂ + H₂O + CO₂

therefore to find the moles used for HCl=(0.917×1)= 0.917moles

to find this vollume= $\frac {mol/l}{Molarity} =$ (0.917 × 1000) ÷ 1.5 = 611. 6 ml

Volume = 611.6 ml