Question

How many mL of 0.565 M HCL are needed to react with 79.2 g of CaCO3?
2HCl+ CaCO3 ---> CaCl2+CO2+ H2O

Answer 1

Answer:

HCl molar mass = 36.45

CaCO₃ molar mass = 100

CaCO3 weight given in the question = 79.2 g

2HCl + CaCO₃ CaCl₂   + CO₂ + H₂O

2 x 36.45 100

? ............... 79.2

= (79.2 x 2 x 36.45) / 100

= 5773.68 / 100

= 57.736 g

The weight of HCl required = 57.736 g

From the Molarity of HCl:

Molarity of HCl = 0.565 (From question)

Molarity = (weight / molar mass) x (1000 / Volume in mL)

0.565 = (57.736 / 36.45) x (1000 / V)

0.565 = (1.58397 x 1000) x V

V = 1583.97 / 0.565

V = 2803.5 mL

Hence the volume of HCl required is 2803.5 mL