jakerossmickelowwdv3 jakerossmickelowwdv3

26-09-2017
Chemistry

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3.0 g aluminum and 6.0 g of bromine react to form AlBr3 2Al+3Br2=2AlBr3 How much product would be produced? How much reagent would remain at the end of this reaction?

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akiyama316 akiyama316

06-10-2017

2Al + 3Br2 -------------> 2AlBr3

3 g Al = 0.11 mol Al.

6 g Br2 = 0.0375 mole bromine (it is diatomic).

moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025.

Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g.
moles of Al left = 0.11 - 0.025 = 0.086

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