Question

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:

Answer 1

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

Answer 2

Answer:

$14g P_2I_4$

Explanation:

we have the balanced equation

$5P_4O_6 +8I_2\longrightarrow 4P_2I_4 +3P_4O_{10}$

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

$P_4O_6= 219.88g/mol\\I_2=253.80 g/mol$

Now we calculate the moles we have of each reagent

$219.88 g P_4O_6\longrightarrow 1 mol\\8.80g P_4O_6 \longrightarrow x\\x= \frac{(8.80)(1)}{219.88}=0.04mol P_4O_6$

$253.80 g I_2\longrightarrow 1 mol\\12.37gI_2 \longrightarrow x\\x= \frac{(12.37)(1)}{253.80}=0.049mol I_2$

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 $P_4O_6$ are needed to react 8 moles of $I_2$    

$8 molI_2 \longrightarrow 5 mol P_4O_6\\0.049molI_2\longrightarrow x \\x=\frac{(0.049)(5)}{8}=0.03 molP_4O_6$

For the 0.049 moles of iodine to react completely we need less than 0.04 mol of $P_4O_6$ this means that the$I_2$ 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 $P_2I_4$

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

$8 mol I_2 \longrightarrow 4 mol P_2I_4\\0.049 molI_2 \longrightarrow x \\x=\frac{(0.049)(4)}{8}=0.0245 molP_2I_4$

With 0.049 mol of iodine the maximum amount of $P_2I_4$ we get is 0.0245 mol

molar mass

$P_2I_4=569.57 g/mol$

with the molar mass of the compound we find the grams

$1 mol P_2I_4 \longrightarrow 569.57 g\\0.0245 mol P_2I_4 \longrightarrow x \\x= \frac{(0.0245)(569.57)}{1}= 14 gP_2I_4$

The maximum mass of$P_2I_4$that can be prepared from 8.80g of P4O6 and 12.37g of iodine is 14 g