If 0.123 g Al is reacted with 1.85 g I2 according to the balanced chemical reaction:
2Al + 3I2 2AlI3 how much excess reagent is left over? (the limiting reagent is Al).
I keep getting .98g but it says I'm not correct.
2Al + 3I2 2AlI3 how much excess reagent is left over? (the limiting reagent is Al).
I keep getting .98g but it says I'm not correct.
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2 Al + 3 I2 → 2 AlI3
(0.123 g Al) / (26.98154 g Al/mol) = 0.0045587 mol Al
(1.85 g I2) / (253.8089 g I2/mol) = 0.0072889 mol I2
0.0045587 mole of Al would react completely with 0.0045587 x (3/2) = 0.00683805 mole of I2, but there is more I2 present than that, so I2 is in excess and Al is the limiting reagent.
((0.0072889 mol I2 initially) - (0.00683805 mol I2 reacted)) x (253.8089 g I2/mol) =
0.114 g I2 left over
(0.123 g Al) / (26.98154 g Al/mol) = 0.0045587 mol Al
(1.85 g I2) / (253.8089 g I2/mol) = 0.0072889 mol I2
0.0045587 mole of Al would react completely with 0.0045587 x (3/2) = 0.00683805 mole of I2, but there is more I2 present than that, so I2 is in excess and Al is the limiting reagent.
((0.0072889 mol I2 initially) - (0.00683805 mol I2 reacted)) x (253.8089 g I2/mol) =
0.114 g I2 left over