posted by Abram on .
How do I solve the following:
500ml .4M aluminum nitrate solution contains ______moles of aluminum ions and _______moles of nitrate ions.
I am helping my fiance and we don't know how to calculate this.
mols = M x L = 0.4M x 0.500L = 0.2mols Al(NO3)3 is what you have.
So you have 0.2 mols Al and 3 x 0.2 = 0.6 mols NO3^-