2C3H6(g) + 2NH3(g) + 3O2(g) -> 2C3H3N(g) + 6H2O(g)
A 150.0 liter steel reactor at 25 ºC is filled with the following partial pressures of reactants: 0.500 MPa C3H6, 0.800 MPa ammonia, and 1.500 MPa oxygen gas.
If the reaction proceeds to 90% yield, and no side reactions take place, what is the final total pressure in the reaction vessel at 400 ºC?
1 answer
This is a bear of a problem but I will get you started.First you must determine the limiting reagent(LR). Start by using PV = nRT and calculating mols of each material. Find the LR from that and determine how much of each product is formed. Then multiply by 0.9 to correct for the 90% yield. Determine how much of the non-limiting reagents remain and add all of the mols together, then convert using PV = nRT to total pressure.
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