Question

Question #67a38

Answer 1

There are 0.200 mol or 19.6 g of sulfuric acid in the solution.

Explanation:

`"Moles of H"_2"SO"_4 = 0.250 color(red)(cancel(color(black)("dm"^3color(white)(l) "solution"))) × ("0.800 mol H"_2"SO"_4)/(1 color(red)(cancel(color(black)("dm"^3color(white)(l) "solution")))) = "0.200 mol H"_2"SO"_4`

`"Mass of H"_2"SO"_4 = 0.200 color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("98.08 g H"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "19.6 g H"_2"SO"_4`