We are going to be synthesizing common alum in lab on thursday. We have prelab questions due by then. The first question is:

What mass of pure aluminum can be dissolved by 13 mL of 1.6M KOH?

I know that there is a 1:1 ratio of Al and KOH when the equation is balanced, but beyond that, I'm not sure exactly how to calculate this.

The other questions I can figure out once I get this mass.

Thank you!

Reaction is
2 Al(s) + 2 KOH(aq) + 6 H2O ---> 2 K[Al(OH)4](aq) + 3 H2(g)
It means two moles are reacting with two moles of KOH
No moles of KOH = molarity X volume in L
= 1.6 X .013L
No of moles of Al = 1.6 X .013L
Multiply it by molar mass of Al to take out the mass