Question

Question #d2de4

Answer 1

You'd need `"1.70 L"` of your `"6.9-M"` `"H"_2"SO"_4` solution to produce that much hydrogen gas.

So, you've got the balanced chemical equation for your reaction

`2Al_((s)) + 3H_2SO_(4(aq)) -> Al_2(SO_4)_(3(aq)) + 3H_2(g)`

A very important thing to notice is that yoiu have a `"1:1"` mole ratio between sulfuric acid and hydrogen gas, i.e. the number of moles of the latter produced will mirror the number of moles of the former that react.

Regardless of how many moles of sulfuric acid react, you'll always produce the same number of moles of hydrogen gas.

This means that, since you know how many moles of hydrogen gas were produced, you can work backwards to see how many moles of sulfuric acid reacted.

Use hydrogen's molar mass to calculate the number of moles produced by the reaction

`"23.7 g hydrogen" * "1mole"/"2.00 g" = "11.75 moles hydrogen"`

Automatically, this will also be the number of moles of sulfuric acid used

`"11.75 moles hydrogen" * "3 moles sulfuric acid"/"3 moles hydrogen" = "11.75 moles sulfuric acid"`

Now all you have to do is use the stock solution's molarity to figure out how much volume would contain this many moles of `H_2SO_4`

`C = n/V => V_(H_2SO_4) = n/C`

`V_(H_2SO_4) = "11.75 moles"/"6.9 mol/L" = "1.703 L"`

Rounded to three sig figs, the answer will be

`V_(H-2SO_4) = "1.70 L"`

SIDE NOTE. I'm not sure about that "minimum amount" in this case, because, assuming aluminium is not a limiting reagent, i.e. you have plenty of it available to react with the sulfuric acid, 23.7 g of hydrogen can only be produced from this much solution.

If you use more solution, you'll essentially make more moles of sulfuric acid available to react `->` you'll produce more moles of hydrogen gas and thus a bigger mass.

Anyway, it's not that important unless some information about how much aluminium you have is given