Question

A quantity 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?

Answer 1

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

Step-by-step 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