Question

Reaction container holds 5.77g of P4 and 5.77g of O2. The following reaction occurs:

Answer 1

Answer:

a. O₂ is limiting reactant

b. 5.68g P₄O₁₀ are produced

c. 5.83g P₄O₆ are left in the reaction container

Explanation:

Based in the first reaction,

P₄ + 3O₂ → P₄O₆

1 mole of P₄ reacts with 3 moles of oxygen

Initial moles of P₄ and O₂ are:

Moles P₄ (Molar mass: 124g/mol):

5.77g P₄ * (1mol / 124g) = 0.0465 moles P₄

Moles O₂ (32g/mol):

5.77g O₂ * (1mol / 32g) = 0.180 moles O₂

For a complete reaction of 0.0465 moles P₄ are required:

0.0465 moles P₄ * (3 moles O₂ / 1 mol P₄) = 0.140 moles of O₂

That means will remain 0.040 moles of O₂ and there are produced 0.0465 moles of P₄O₆

For the second reaction:

P₄O₆ + 2O₂ → P₄O₁₀

2 moles of oxygen reacts per mole of P₄O₆

For a complete reaction of P₄O₆ are required:

0.0465 moles P₄O₆ * (2 moles O₂ / 1mol P₄O₆) = 0.093 moles of O₂. As there are just 0.040 moles of O₂,

a. O₂ is limiting reactant

b. There are produced:

0.040 moles O₂ * (1 mole P₄O₁₀ / 2 moles O₂) = 0.020 moles P₄O₁₀

In mass (Molar mass: 284g/mol):

0.020 moles * (284g / mol) =

5.68g P₄O₁₀ are produced

c. Also, 0.020 moles P₄O₆ are reacting and will remain:

0.0465 mol - 0.020 mol = 0.0265 moles P₄O₆

In mass (Molar mass: 220g/mol):

0.0265 moles P₄O₆ * (220g / mol) =

5.83g P₄O₆ is left in the reaction container