A piece of aluminum foil 1.00 cm square and .550 mm thick is allowed to react with bromine to form aluminum bromide.
a) how many moles of aluminum were used? (density of aluminum = 2.699g/cm^3)
b) how many grams of aluminum bromide form assuming that the aluminum reacts completely?
Could you show your work please? Thank You
1 Answer

a)
work out the volume of Al in cm^3
Volume = l x b x h
= 100 cm x 1.00 cm x 0.0550 cm
= 0.0550 cm^3
then work out the mass
density = mass / Volume
Therefore mass = density x volume
= 2.699 g/cm^3 x 0.0550 cm^3
= 0.148445 g
moles = mass / molar mass
= 0.148445 g / 26.98 g/mol
= 0.00550 mol
b)
write a balanced equation
2Al + 3Br2 > 2AlBr3
2 moles Al react to form 2 moles AlBr3
A 1 : 1 reaction ratio
Therefore maximum moles AlBr3 possible = moles Al used
= 0.00550 mol AlBr3 possible
mass = molar mass x moles
mass AlBr3 = 266.68 g/mol x 0.00550 mol
= 1.47 g (3 sig figs)