Question

A sample of hydrochloric acid contains 20% HCL by weight how much of this acid in grams is required for complete reaction with 50 gram of calcium carbonate

Answer 1

CaCO3 + 2HCl ---------> CaCl2 + H2CO3

So we can say one mole of CaCO3(100 g) require two moles of 100% HCl for complete neutralisation.

So 50 g will require one mole of 100% HCl

For 20% HCl,

Let x be the required amount,

20/100 × x = 35.5

× = 35.5 × 100/20

= 35.5 × 5

= 177.5 g