Question

Question #bfee9

Answer 1

`"1.5 L H"_2"SO"_4`

Explanation:

Since you are given grams of `H_2`, it will be easiest to work backwards using the stoichiometry.

Therefore:

`"15 g H"_2 xx ("1 mol H"_2)/("2.016 g H"_2) xx ("3 mol H"_2"SO"_4)/("3 mol H"_2) = "7.44 mol H"_2"SO"_4`

You could go further in the analysis, but you are given the molarity of the sulfuric acid. That should give you some guidance in the direction to solve this problem.

Therefore, using the equation

`"M"= ("mol")/("L")`

we can plug in the number and solve for the volume.

`"5.0 " "mol"/"L" = ("7.44 mol")/("?")`

`? = ("7.44 mol H"_2"SO"_4)/(5.0 "mol"/"L")`

`? = "1.5 L H"_2"SO"_4`

I hope that helps!