Hi,
To solve the question, first of all we will find out the no. of moles of H2SO4 in 19 g of sulfuric acid.
As we know:
No . of moles = Mass/ Molar mass
No. of moles= 19 g/98.08
g
No. of moles= 0.1937
Now we know the no of moles of H2SO4 that will react with 2LiOH. We also know the molar equivalence of H2SO4 , and 2LiOH that will react.
So, the water that will be produced will be 2H2O and 1 Li2SO4 when H2SO4 that will react with 2LiOH.
0.1937 x 2x 18.01
=6.977
=6.98
Therefore, approximately 6.98 grams of water will be produced from 19 g of sulfuric acid.
Hope it helps!
