Question

Suppose a 500. mL flask is filled with 0.60 mol of CH, 0.70 mol of CO and 0.10 mol of H,. This reaction becomes possible: CH4(8)+H20(g)Co (g)+3H2(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of H2O. You can leave out the M symbol for molarity. сн, н,о н, Co initial ? change equilibrium

Answer 1

CH4 + H₂O - co + 3H₂(g) (9) g) (g) *9) 0.60 mol = 2 M 0.54 [Chhat] mol = 0.60 m) = 1,2M [co] = mol = on the m= 1.4m [12] = 05 = $ = 0.2% 0.7 0.5 M= -4M + CHA 102 H2O - co + 3H2 Initial ly 0.2 La _3a Change ta + 114-a 0.2-32 u lo2ta ICH4 l H2O - col H2 CHEE 1.4 0:2 Initial 32 Change a 14-a 10:2-3a. E Equilionum 12ta