2 Al + 3 I2 →2 AlI3
Mm I2 = 253.8 g/mol
AlI3 = 407.70 g/mol
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2 Al + 3 I2 →2 AlI3
Mm I2 = 253.8 g/mol
AlI3 = 407.70 g/mol
g AlI3 = 25 g I2 x 1 mol I2 x 2 mol AlI3 x 407.70 g AlI3```````````` `````````````` ``````````````````
253.8 g I2 3 mol I2 1 mo AlI3
g AlI3 = 26.77