Answer:
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
