How many moles of aluminum were used? (The density of aluminum is 2.699g/cm3 .)?
A piece of aluminum foil measuring 1.15 cm on each side and 0.570 cm thick is allowed to react with bromine to form aluminum bromide. How many moles of aluminum were used in this reaction? Given that the density of aluminum is 2.699 g/cm³, how many grams of aluminum bromide will be produced, assuming that the aluminum reacts completely?
3 Answers
2 Al + 3Br2 >> 2 AlBr3
Volume Al = 1.15 x 0.570 = 0.656 cm^3
Mass = 0.656 x 2.699 = 1.77 g
Moles Al = 1.77 g / 26.9815 g/mol =0.0656
the ratio between Al and AlBr3 is 2 : 2 ( or 1 : 1)
moles AlBr3 produced = 0.0656
Mass AlBr3 = 0.0656 mol x 266.6935 g/mol = 17.5 g
1,15cm square is Not the same as 1,15 square cm. 1,15 cm square means a square with a side of length 1,15 cm. To avoid this type of confusion use the scientific notation cm^2.
So the answer will depend on how you translate this part of the problem. Either translation could be correct.
So area of Al foil = 1,15 X 1,15 cm^2 …. Agreed? Then volume
= 1,15 X 1,15 X 0,570 cm^3
= 0,7538 cm^3
Mass of Al = 0,7538 X 2,699 g
= 2,0346 g ( A M of Al = 26,98 )
Moles Al = 2,0346/ 26,98 = 0,0754 Moles
AlBr3 is formed so 3 X 79,9 g of Br combine with 26,98 g Al to form 266,68 g AlBr3.
So mass of AlBr3 = ( 2,0346/26,98 ) X 266,68 g = 20,11 g
Al used = 0,0754 Moles
AlBr3 formed = 20,11 grams
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