Solid white phosphorus, P_4, ignites spontaneously in air at temperatures above

Question

Solid white phosphorus, P_4, ignites spontaneously in air at temperatures above 50 degree C to produce solid diphosporus pentoxide. For an experiment a scientist needs to produce 275 kg of diphosphorus pentoxide. If the reaction proceeds with an 82% yield, what is the volume of oxygen gas (in liters) that is required for the process? MM [P_4) = _____g/mol MM(O_2) = ____g/mol MM(P_2O_5) = g/mol The balanced chemical equation that represents this reaction: A. Solve the problem: B. Assess: Does your answer make sense? It is important to get into the habit of not just performing a series of math steps. Take a moment to assess your work. Practice now by providing a justification for your answer in part A. You can do this in a few different ways. One way is to work out the units; the other way is to try to estimate the value to see if you get the correct order of magnitude.

Explanation / Answer

The reaction between white phosphorous and Oxygen is

P4+ 5O2-----à2P2O5

Molar masses : P4= 31*4= 124, O2=32 and P2O5= 142

Moles of P2O5 to be produced= mass of P2O5/molar mass= 275*1000/142=1937 gmoles

As per the reaction, 2 moles of P2O5 requires 5 moles of oxygen ( when completely converted).

1937 gmoles of P2O5 requires 1937*5/2 moles of oxygen = 4842.5 moles.

But the reaction is 82% complete. Hence moles of oxygen actually required= 4842.5/0.82=5905.5 moles.

1 mole of any gas at STP occupies 22.4 L

5905.5 moles of gas occupies 5905.5*22.4 L=132283 L=132 m3. So the required volume of reactor will be very huges and does not look practicable.

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