Answer:
You must add 13.86g of I₂
Step-by-step 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