DATA

Mass of P₄ = 5.77g
Mass of 0₂ = 5.77g

Equations PO₄ + 30₂ → P₄O₆
P₄O₆ + 20₂ → P₄O₁₀
Asked
i. Mole of P₄O₁₀ produced
ii. Mass of P₄O₁₀ produced

To find the mole of P₄,
Mole of P₄ = mass of P₄ / Molar mass of P₄
= 5.77g / 124g/mole
= 0.0465 mole

Mole of 0₂ = mass of 0₂ / Mole of 0₂
= 5.77g / 32g/mole
= 0.1803125 mole

Mole ratios for P₄ = 0.0465 / 1
= 0.0465
For 0₂ = 0.803125 / 3
= 0.060104
The limiting reagent was oxygen

i/ To find mole of P₄O₁₀

mole equivalent between O₂ and P₄O₁₀
2 miles of O₂ Ξ 1 mole of P₄O₁₀ 0.1803 mole Ξ?
Mole of O2 = (0.1803 mole × 1 mole) / 2 mole
= 0.09015 mole
0.09015 mole of P₄O₁₀ produced

ii/ To find the mass of P₄O₁₀

From;
Mole of P₄O₁₀ = 0.09015 mole
Molar mass of P₄O₁₀ = 284
Mass = mole × molar mass
= 0.0915 mole × 284g/mole
= 256.026g
256.026g of P₄O₁₀ produced