Question

Consider the reaction of phospene, ph3 with oxygen: 4ph3 + 8o2 ->p4o10 + 6h2o. Suppose a 100 kg mixture of 50% ph3 and 50% o2 by mass enters a reactor in a single stream, and single exit stream contains 25% o2 by mass. Determine mass composition of exit stream.

Answer 1

Given data: 100 Kg of reaction mixture with 50% PH₃ and 50%O₂ ,

Mass of PH₃ = 0.5*100 = 50Kg

Moles of PH₃ = Mass/molar mass = 50/34 = 1.47 Kmoles

Mass of O₂ = 0.5 *100 = 50 Kg

Moles of O₂ = Mass /Molae mass = 50/32 = 1.5625 kmoles

Given reaction,

4PH₃ + 8O₂ P₄O₁₀ + 6H₂O

For every moles of Phosphine we need 8 moles of O₂. But our reactant mixture has 1.47 Kmol of PH₃ and 1.5625 Kmol of O₂ , thus O₂ is the limiting reactant.

Let x be the conversion ,

Moles of O₂ converted = 1.5625x

Moles of O₂ in exit stream = 1.52625*(1-x)

Moles of PH₃ converted = Moles of O₂ converted * 4/8 = 1.5625x/2

Then PH₃ exit stream = 1.47 - 1.5265x/2 = 1.47 - 0.78125x

Moles of P₄O₁₀ formed = Moles of O₂ converted/8 ( from stoichiometry)

= 1.5625x/8 = 0.1953125x

Moles of H₂O formed = Moles of O₂ converted * 6/8 = 1.171875x

Mass of O₂ in exit = Moles of O₂ in exit * Molar mass = 1.52625*(1-x)* 32 = 50(1-x) = 50 - 50x

Mass of PH₃ in exit = Moles of PH₃ in exit * Molar mass = (1.47 - 0.78125x)*34 = 50 - 26.5625x

Mass of  P₄O₁₀ in exit = Moles of  P₄O₁₀in exit * Molar mass = 0.1953125x * 284 = 55.46875x

Mass of H₂O in exit = Moles of H₂O in exit * Molar mass = 1.171875x * 18 = 21.09375x

Given that mass percent of O₂ inexit stream = 25% , in terms of mass fraction, = 0.25

Mass of O₂/ Total mass of exit stream = 0.25

50(1-x) / [50 - 50x +  50 - 26.5625x + 55.46875x + 21.09375x ] = 0.25 ,

but we know that total mass exit = total mass at inlet = 100 Kg , solving this gives

x = 0.5

Mass of O₂ = 50 - 50x = 25 Kg

Mass of PH₃ = 50 - 26.5625x = 36.718 Kg

Mass of P₄O₁₀= 55.46875x = 27.734 Kg

Mass of H₂O = 21.09375x = 10.546 Kg

Mass percentage of PH₃ = (Mass of PH₃ in Exit / Total mass )*100 = 34.718/100 * 100 = 34.718%

Mass percentage of P₄O₁₀ = (Mass of P₄O₁₀ in Exit / Total mass )*100 = 27.734/100 * 100 = 27.734%

Mass percentage of H₂O= (Mass of H₂O in Exit / Total mass )*100 = 10.546/100 * 100 = 10.546 %