What is the maximum mass of P2I4 that can be prepared from 8.80g of P4O6 and 12.37g of iodine according to the reaction:

5P4O6 + 8I2 - - - > 4P2I4 + 3P4O10

The balanced chemical reaction would be as follows:

5P4O6 + 8I2 - - - > 4P2I4 + 3P4O10

We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:

8.80 g P4O6 (1 mol / 219.88 g) = 0.04 mol P4O6

12.37 g I2 (1 mol / 253.809 g) = 0.05 mol I2

Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:

0.05 mol I2 (4 mol P2I4 / 8 mol I2) (569.57 g / 1 mol) = 14.24 g P2I4