How many grams of Br2Br2 are needed to form 74.1 g74.1 g of AlBr3AlBr3? 2Al(s)+3Br2(l)⟶2AlBr3(s)2Al(s)+3⁢Br 2⁢(l)⟶2⁢AlBr3⁢(s)
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grams AlBr3⟶⟶⟶grams Br2grams AlBr3⟶⟶⟶grams Br2
Answer Bank
moles Br2Br2
moles AlAl
grams Br2Br2
grams AlBr3AlBr3
grams AlAl
moles AlBr3AlBr3

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