Consider the reaction 5P4O6 + 8I2 4P2I4 + 3P4O10. How many grams of iodine must be added to 7.15 g of P4O6 to have a 5.00 % excess of iodine?

You must add 13.86g of I₂

Explanation:

In the reaction:

5 P₄O₆ + 8I₂ → 4P₂I₄ + 3P₄O₁₀

5 moles of P₄O₆ reacts with 8 moles of I₂.

First, we need to determine how many moles of I₂ reacts completely with 7.15g of P₄O₆ (Molar mass: 219.88g/mol, thus:

7.15g P₄O₆ * (1mol / 219.88g) = 0.0325 moles P₄O₆.

Using the chemical reaction, the moles of I₂ that reacts are:

0.0325 moles P₄O₆ * ( 8 moles I₂ / 5 moles P₄O₆) = 0.0520 moles of I₂ react completely.

As molar mass of I₂ is 253.81g/mol, the mass of I₂ that reacts completely is:

0.0520 moles of I₂ * (253.81g / mol) =

13.2g of I₂.

As you must add 5.00% of excess:

13.2g + 5%13.2g =

You must add 13.86g of I₂