50.0 g of N2O4 is introduced into an evacuated 2.00 Litre vessel and allowed to come to equilibrium with its decomposition product, N2O4(g) = 2 NO2(g). For this reaction Kc=0.133. Once the system has reached equilibrium, 5.00 g of NO2 additional is injected in to the vessel, and the system is allowed to equilibrate once again. Calculate the mass of NO2 in the final equilibrium mixture.
To view this in proper format, cut and paste it into a simple text editor like word pad.... I put in multiple spaces, but when I post this, all the spacing gets messed up... anyway... I will try using multiple (....) for spacing instead.
First I convert grams of N2O4 to moles... ALWAYS do that!
Moles N2O4 = 50.0 g/92.01 g/mol = 0.543 mol
Initial concentration of N2O4 = 0.543 mol/2.00 L = 0.272 M
And grams of NO2 to ...
The solution provides a detailed, step-by-step calculation for the mass of a molecule in a final equilibrium mixture of a decomposed product.