What's the limiting reagent? See discussion below.
...........A + 3B ==> AB3
initial..0.6...1.2.....0
change..-0.4..-1.2....+0.4
equil....0.2....0.....0.4
0.6 atm A will form 0.6 atm AB3 but
1.2 atm B will form 1.2 x (1/3) = 0.4 atm AB3; therefore, B is the limiting reagent.
Total pressure is 0.2 + 0.4 = ? atm and partial pressures are given in the equilibrium line.
You can work this by assuming some T and some V, solve for moles and work it with moles. Then use PV = nRT and solve for pressures of each but it takes much longer. I get the same answer either way.
Consider the following reaction in a closed reaction flask. If 0.600 atm of gas A is allowed to react with 1.20 atm of gas B and the reaction goes to completion at constant temperature and volume, what is the total pressure in the reaction flask at the end of the reaction? What are the partial pressures of the gases in the flask?
A(g) + 3B(g) → AB3(g)
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