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

Explanation:

we have the balanced equation

First we must determine which is the limit reagent

The limit reagent is one that runs out first and when it runs out, the reaction is over

It is the one that determines how much product is going to be produced

molar masses

Now we calculate the moles we have of each reagent

To calculate the limit reagent we use the stoichiometric coefficients of the reagents and apply a rule of three

From the ratio of stoichiometric coefficients, we know that 5 mol of are needed to react 8 moles of    

For the 0.049 moles of iodine to react completely we need less than 0.04 mol of this means that the is going to run out first therefore it is our limit reagent

From the ratio of stoichiometric coefficients we know that 8 moles of iodine are needed to produce 4 moles

To calculate how much will be produced with 0.049 mol of iodine we apply a rule of three

With 0.049 mol of iodine the maximum amount of we get is 0.0245 mol

molar mass

with the molar mass of the compound we find the grams

The maximum mass ofthat can be prepared from 8.80g of P4O6 and 12.37g of iodine is 14 g

16

explanation:

the marks go by 4s