A 8.50-g sample of solid AlCl3·6H2O was heated such that the water turned to steam and was driven off. Assuming ideal behavior, what volume would that steam occupy at 1.00 atm and 100.0 °C?

Answer 1

AlCl3.6H2O ==> AlCl3 + 6H2O
mols AlCl3 = 8.5g/molar mass AlCl3 = approx 0.06 but you need a more accurate number.
Convert mols AlCl3 mols H2O using the coefficients in the balanced equation. That's about 6 x 0.06 = about 0.36.
Then plug that into for n in PV = nRT and solve for V in liters. Don't forget T must be in kelvin.