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

Question

A piece of aluminum foil 1.15cm square and 0.570cm thick is allowed to react with bromine to form aluminum bromide.How many moles of aluminum were used? (The density of aluminum is 2.699g/cm3 .) How many grams of aluminum bromide form, assuming that the aluminum reacts completely?

Prof. Carmelo Hilpert · Accepted Answer

2 Al + 3Br2 >> 2 AlBr3Volume Al = 1.15 x 0.570 = 0.656 cm^3Mass = 0.656 x 2.699 = 1.77 gMoles Al = 1.77 g / 26.9815 g/mol =0.0656the ratio between Al and AlBr3 is 2 : 2 ( or 1 : 1)moles AlBr3 produced = 0.0656Mass AlBr3 = 0.0656 mol x 266.6935 g/mol = 17.5 g

Prof. Carmelo Hilpert · Answer

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^3Mass 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 MolesAlBr3 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 gAl used = 0,0754 MolesAlBr3 formed = 20,11 grams

Prof. Carmelo Hilpert · Answer

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

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