MCQ Detailed View

The reaction between phosphorus and oxygen to form phosphorus pentoxide is a classic stoichiometry problem that tests understanding of limiting reagents. The balanced equation is:

P4+5O2→P4O10P₄ + 5O₂ \rightarrow P₄O₁₀P4+5O2→P4O10

From the equation, one mole of P4P₄P4 reacts with five moles of oxygen to yield one mole of phosphorus pentoxide.

First, we calculate the moles of reactants. The molar mass of P4P₄P4 is 124 g/mol, so 1.33 g corresponds to $1.33/124\approx 0.0107mol$. For oxygen, with a molar mass of 32 g/mol, 5.07 g corresponds to $5.07/32\approx 0.1587mol$.

Next, we identify the limiting reagent. For 0.0107 mol of P4P₄P4, the required oxygen is $0.0107\times 5=0.0535mol$. Since the available oxygen is 0.1587 mol (greater than required), oxygen is in excess, and phosphorus is the limiting reagent. This means the reaction yield is governed by the amount of P4P₄P4.

Now, we calculate the product mass. The reaction shows a 1:1 mole ratio between P4P₄P4 and P4O10P₄O₁₀P4O10. Hence, 0.0107 mol of P4P₄P4 yields 0.0107 mol of P4O10P₄O₁₀P4O10. With a molar mass of 284 g/mol, the product mass is $0.0107\times 284\approx 3.05g$.

Thus, the mass of phosphorus pentoxide formed is 3.05 g.

This problem emphasizes the importance of carefully applying stoichiometric ratios and identifying the limiting reagent, as it controls the amount of product formed even when other reactants are present in excess.