Aquantity of liquid methanol, ch3oh, is introduced intoa rigid 3.00-l vessel, the vessel is sealed, and the temperature israised to 500k. at this temperature, the methanol vaporizes anddecomposes according to the reactionch3oh(g) a. 147 g b. 74.3 g c. 33.9 g d. 49.0 g e. 24.8 gco(g)+ 2 h2(g), kc= 6.90�10^-2. if the concentration of h2 in the equilibrium mixture is0.426 m, what mass of methanol was initially introduced into thevessel?

74,3 grams is the mass of methanol that was initially introduced into the vessel

Explanation:

                         CH3OH(g)        >          1 CO(g)    +   2 H2(g)

initial:             Idk what i have (X)                    -                    -

react:                initial - react                        react              react

equilibrium:    A concentration                  A conc.         0.426 M

                         in equilibrum                      in eq      

See that in equilibrium are formed 2 moles of H2 with this concentration, 0,426M and you form 1 mol of CO, so the moles that are formed of H2 are the double of CO, because stoychiometry.

                            CH3OH(g)        >          1 CO(g)    +   2 H2(g)

initial:                 Idk what i have (X)                   -                    -

react:                    initial - react                     react              react

equilibrium:        A conc. in eq                     0,213 M         0.426 M

So we have concentrations of products in equilibrium, and we have Kc, now we can find concentration of reactants in equilibrium

Kc = ([CO] . [H2]*2) / [CH3OH]

6,9X10*-2 = (0,213 . 0,426*2) / [CH3OH]

[CH3OH] = (0,213 . 0,426*2) / 6,9X10*-2

[CH3OH] = 0,560 M

You know that 1 mol which reacted has this concentration in equilibrium, 0,213 M so I can know what's my initial concentration of reactive

Initial - react = equilibrium

initial - 0,213M = 0,560M ---> Initial = 0,773 M

0,773M is initial concentration of CH3OH, but this is molarity, (moles in 1L). My volume is 3 L so

1 L 0,773 moles

3 L  3L . 0,773moles = 2,319 (the moles that i used)

Molar mass CH3OH = 32.06 g/m

Moles . molar mass = grams ---> 2,319 moles . 32,06 g/m  = 74,3 g                                                

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