The equilibrium constant, Kc, for the following reaction is 3.61×10-4 at 426 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently lar
ge sample of PCl5(g) is introduced into an evacuated vessel at 426 K, the equilibrium concentration of Cl2(g) is found to be 0.327 M. Calculate the concentration of PCl5 in the equilibrium mixture.
1 answer:
Answer:
the concentration of PCl5 in the equilibrium mixture = 296.20M
Explanation:
The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.
The detaied steps is as shown in the attached file.
You might be interested in
Answer:
Explanation:
Hello,
For the given chemical reaction:
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
Finally, we compute the percent yield with the obtained 2.10 g:
Best regards.
Answer:
The volume of hydrogen gas produced at STP is 4.90 liters.
Explanation:
Moles of aluminium =
According to reaction , 2 moles of aluminium gives 3 moles of hydrogen gas.
Then 0.1333 moles of aluminium will give:
of hydrogen gas
Volume of 0.2 moles of hydrogen gas at STP = V
Temperature at STP = T = 298.15 K
Pressure at STP = P = 1 atm
n = 0.2 mol
PV = nRT (Ideal gas equation)
The volume of hydrogen gas produced at STP is 4.90 liters.
n' = 4 ( solution is attached below )
