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OpenStudy (anonymous):
How many grams of Br2 are needed to react completely with 19.6 grams of Al?Al + Br2 AlBr3Be sure to balance the equation. Answers: 38.7 grams 43.6 grams 87.1 grams 174 grams
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OpenStudy (jfraser):
put together the balanced equation, first
OpenStudy (anonymous):
You will do a conversion into the mol/mol ratio. I don't know if you need to show your work, but here it is: 19.6g Al (1mol Al/26.982g Al)(3mol Br2/2mol Al) (159.818g Br2/1mol Br2) = 174.14g Br2 So it looks as though D is your answer
OpenStudy (anonymous):
Also the balanced equation looks like this (providing your numbers in the original equation are correct) 2Al + 3Br2 --> 2AlBr3
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