A piece of aluminum foil 0.550 mm thick and 1.00 cm square is allowed to react with bromine to form aluminum bromide?a) How many mole of aluminum were used? ( The density of aluminum is 2.699 g/cm3.)I keep getting the same answer which I know is wrong. I used a formula to get the volume and then the mass to get the moles, but maybe this is wrong. Could someone explain the process. I got 4.32 x 10-1 mol Al, but the answer is 5.50 x 10-3 mol Al.
Okay. If you have .55mm thick, 1cm square of Al, you you have 55 cubic mm of Al. 55 cubic mm is .055 cubic centimeters of Al. .055 * denisty of Al (2.699 g/cm3) gives you .148 grams of Al. Al molecule weight is 27g/mole. So .148 g * (1 mole / 27 grams) = .00548 moles of Aluminum. That rounds to about 5.50 x 10-3 moles. I should also add that A) Remember when you are converting cubic units (such as mm3 into cm3), the conversion factor should also be cubed (so 1 cubic mm = 10^3 cubic cm). Also, not to be nit picky, but this question is dumb because aluminum foil would also have a good portion of aluminum oxide coating it...but I wouldn't bring that up to the teacher if I were you. :-)
0.550mm = 0.055cm 0.055 x 1 x 1 = 0.055cc x 2.699g/cc = 0.1484g Al. Mol.mass = 27 g/mole 0.1484 ÷ 27 = 0.0055 moles. (5.5 x10-?)
volume= 0.0550 cm x 1.00 cm^2= 0.0550 cm^3 gms Al= 2.699 g/cc x [vol] 0.0550 cc = 0.148 g moles Al = [grams]/[mole wt] mol wt=26.98 so 2.699[g/cc] x 0.0550[cc] / 26.98 =0.00550 or 5.50x 10^-3 to 3 sig fig units cancel and gram-moles is final unit.
