Question

The mass of 80% pure H2SO4 required to completely neutralise 106g of Na2Co3

Answer 1

Na2SO3 + H2SO4 → Na2SO4 + SO2 + H2O

106 gm
mole = 106/126
         = 0.84

by stoichiometry coefficient mole mole of H2SO4 
= 0.84 = 80% of (wt/mw)
   0.84 = 0.8 x (wt/98)

so weight or mass = (98 x 0.84) / 0.8 
                              = 103 gm

i hope it will help you
regards

Answer 2

Answer: 122.5 g

Explanation:

$H_2SO4(aq)+Na_2CO_3(s)\rightarrow Na_2SO_4(aq)+H_2O+CO_2(g)$

According to the balanced chemical equation, 1 mole of $H_2SO_4$ reacts with 1 mole of $Na_2CO_3$.

Mass of 1 mole of $H_2SO_4$ = 98g

Mass of 1 mole of $Na_2CO_3$ = 106 g

Thus 106 g of  $Na_2CO_3$ reacts with 98 g of $H_2SO_4$ for complete neutralization.

But the percentage purity of $H_2SO_4$ is 80 %, thus mass of $H_2SO_4$ required will be:

$\frac{80}{100}\times x= 98$

$x= 122.5 g$

Thus mass of 80% pure $H_2SO_4$ required to completely neutralise 106g of $Na_2CO_3$ is 122.5 g