[narutodemonkill]

When aluminium metal is added to suplhuric acid hydrogen gas and aluminum sulphate is made. what volume of hydrogen gas will evolve if we use 15.00 ml of .50 M solution of sulphuric acid and add 0.0712 g of aluminum? Assume that this reaction is only 90.0% efficient and the the experiment was performed at 1.00 atm and 25.0 degrees Celsius.

Kind of stuck..
I got the equation 2Al(s) + 3H2so4(aq) --> 3H2(g) +Al2(so4)3(aq)
What i would do is find mols of sulphuric acid and mols of aluminum and figure out which one is limiting reagent than find the mols of hydrogen gas but how would you find volume of hydrogen gas?
c=n/v n/c=volume but we do not know concentration of hydrogen gas?

Any ideas?
because it is 90% efficient w.e answer you would get you divide it by .90 right.

I figured it out you use pv=nrt = volume the problem is in one example my teacher did in which reaction was 31% efficient she divided the final volume by .31
do I do the same here? i.e divide final volume by .90 and why?

[UG]

Any ideas?
because it is 90% efficient w.e answer you would get you divide it by .90 right.

I figured it out you use pv=nrt = volume the problem is in one example my teacher did in which reaction was 31% efficient she divided the final volume by .31
do I do the same here? i.e divide final volume by .90 and why?

Divide? You sure she didn't multiply by 0.31? Dividing the final volume would give you an even larger volume.
But your method looks right, V(H₂)=nRT/P. That's no problem