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If 0.333 g Al is reacted with .501 g I2 according to the balanced chemical reaction:
2Al + 3I2 2AlI3
a) what is the limiting reagent
b) how much excess reagent left over?
1. Place an arrow at the appropriate spot so you can tell the reactants from the products.
2. Convert 0.333 g Al to moles. moles = g/molar mass. Convert 0.501 g I2 to moles.
3a. Using the coefficients in the balanced equation, convert moles Al to moles AlI3.
3b. Using the same procedure, convert moles I2 to moles AlI3.
3c. The smaller number from 3a versus 3b is the limiting reagent AND is the correct value for the number of moles AlI3 formed.
For the second part, using the limiting reagent, go through the same kind of procedure to determine how much of the OTHER reagent is needed to react with the limiting reagent, then subtract that amount from the initial amount to determine how much remains un-reacted.
Post your work if you get stuck.