Question

find the weight of 90 % pure HCL required to react completely with 5g of 80% pure Ca

Answer 1

Amount of CaCO3 required is 0.94gm.

Given:-

Volume of HCl = 25 ml

Molarity of HCl = 0.75 M

Atomic Mass of Hydrogen = 1g

Atomic Mass of Chlorine = 35.5g

Molecular Mass of HCl = 1 + 35.5

= 36.5g

By Avogadro’s Law,

1 mole of HCl→ 36.5g of HCl

So 0.75 moles of HCl → X g of HCl

X = 0.75×36.5

X = 27.35 g

Thus, 1000 mL of solution contains HCl = 27.375g

1000ml of HCl → 27.357g of HCl

So 25 ml of HCl → X g of HCl

X = [25×27.375]/1000

= 684.375/1000

=0.68438 g

We observe that

2 moles of HCl reacts with 1 mole of CaCO3

73 g of HCl reacts with 100g of CaCO3

0.68438g of HCl reacts with X g of CaCO3

X = [0.68483×100]/73

= 68.483/73

= 0.93812 g

So amount of CaCO3 required is 0.94g.