Stoichiometry: Calculate the mass of P₄O₁₀ formed from given reactants. Reaction: P₄ + 5O₂ → P₄O₁₀ Given 1.33 g of P₄ and 5.07 g of O₂ react completely under standard conditions. Find the mass of P₄O₁₀ produced (assume complete reaction and pure reagents). Choose the correct value.

Given

• Reaction: P₄ + 5O₂ → P₄O₁₀
• m(P₄) = 1.33 g; M(P₄) = 4×31 = 124 g/mol
• m(O₂) = 5.07 g; M(O₂) = 32 g/mol
• M(P₄O₁₀) = 4×31 + 10×16 = 124 + 160 = 284 g/mol

Approach
Find the limiting reagent, then compute product moles and mass.

Step-by-step
n(P₄) = 1.33/124 = 0.010726 moln(O₂) = 5.07/32 = 0.15844 molO₂ needed = 5×0.010726 = 0.05363 mol (available 0.15844 mol → excess)P₄ is limiting, so n(P₄O₁₀) = n(P₄) = 0.010726 molm(P₄O₁₀) = 0.010726 × 284 = 3.046 g

Verification
Significant figures from 1.33 and 5.07 → ~3.05 g.

Final Answer
3.046 g (≈ 3.05 g).