Question

What is the maximum mass of P₂I₄ that can be prepared from 8.19 g of P₄O₆ and

11.21 g

of iodine according to the reaction

5P₄O₆ + 8I₂

→

4P₂I₄ + 3P₄O₁₀

Answer 1

The balanced equation is

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

5 x 220 8x 254 4x569.6g/mol  

First we need to find the limiting reagent

8.19/220= 0.0372 11.21/254=0.0441

0.0372/5=0.00745 0.0441/8=0.0055

As the ratio of I2 is less than that of P4O6, I2 is the limiting reagent

Let us calculate the mass of P2I4 that can be formed.

8x254g of I2 gives  4x569.6g of P2I4 on reaction

thus 11.21 g of I2 gives = 11.21 gx 4x569.6g/8x254 g

= 12.57 g of P2I4

12.57 g of P2I4 is the maximum that can be btained from the reaction.