Question #67a38

1 Answer
Ernest Z.
Feb 28, 2017

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#

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