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
<em>1 mole of P₄ reacts with 3 moles of oxygen</em>
Initial moles of P₄ and O₂ are:
<em>Moles P₄ (Molar mass: 124g/mol): </em>
5.77g P₄ * (1mol / 124g) = 0.0465 moles P₄
<em>Moles O₂ (32g/mol):</em>
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
<em>That means will remain 0.040 moles of O₂ and there are produced 0.0465 moles of P₄O₆</em>
<em />
For the second reaction:
P₄O₆ + 2O₂ → P₄O₁₀
<em>2 moles of oxygen reacts per mole of P₄O₆</em>
For a complete reaction of P₄O₆ are required:
<em />
0.0465 moles P₄O₆ * (2 moles O₂ / 1mol P₄O₆) = 0.093 moles of O₂. As there are just 0.040 moles of O₂,
<h3>
a. O₂ is limiting reactant</h3><h3 />
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) =
<h3>5.68g P₄O₁₀ are produced</h3>
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) =
<h3>5.83g P₄O₆ is left in the reaction container</h3>
