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A

Anonymous

Nov 13, 2024

Chemistry Homework Problem?

A piece of aluminum foil measuring 1.00 cm on each side and 0.550 mm thick is allowed to react with bromine to form aluminum bromide.

a) How many moles of aluminum were used? (The density of aluminum is 2.699 g/cm³.)

b) Assuming that the aluminum reacts completely, how many grams of aluminum bromide are formed?

Could you please show your work? Thank you.

1 Answers

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)

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