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From your question:
[H2SO4] = 6 M
Mass H2 = 25 g
- First - be careful writing formula and symbols
- Second - I am unclear as to the number of significant figures, I will use 3
- By amount (which is the SI unit of measurement with a unit of mole) I infer you mean volume
[H2SO4] = 6.00 M
Mass H2 = 25.0 g
Balanced equation will give mole ratio:
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
Mole ratio, H2SO4:H2 = 1:1
Moles H2 required = 25.0 g / 2.00 g mol-1
=12.5 mol
= 12.5 mol H2SO4
Concentration = moles/volume
6.00 M =12.5 mol / volume
Volume = 12.5 / 6.00
= 2.08 dm3 or 2.08 L
Thanks Much:) this place makes writing equations difficult with the syntax rules. You got my intention right and that is the same answer I ended up figuring out so thanks again.
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