Question

The relative molecular mass of calcium carbonate is 100.
What is the minimum volume of 2.0 M hydrochloric acid that would be needed to completely react with 2.0 g of calcium carbonate?

options are
A) 20 cm^3
B) 10 cm^3
C) 5 cm^3
D) 30 cm^3
E) 40 cm^3


I wrote down what I believe is the balanced reaction: CaCO3 + 2HCl -> H2CO3 + CaCl2

but I'm not sure how to solve it

Answer 1

moles of CaCO3=0.02
by comparing moles of HCl=0.04
mass of Hcl=0.04*36.5=1.46g

we know that the molarity of Hcl solution is 2
and its percentage 7
SO
7g of HCl present in Hcl solution=100g
1g of HCl present in Hcl solution=100/7
1.46g of Hcl present in HCl solution=100/7*1.46=20.85

d=m/v

1.03g/cm3=20.85g/V

so it will be 20.24cm3

Answer 2

g HCl is OK but you don't need that. You know you need 0.04 mols HCl; therefore,
M HCl = mols HCl/L HCl
You know M and mols, solve for L.

That 7 stuff of percentage is hog wash.