Question

the mass of 80% pure H2SO4 required to completely neutralise 106g of Na2CO3 is​

Answer 1

Answer:

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.....