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A quantity of liquid methanol, CH3OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500 K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH3OH(g)CO(g) + 2 H2(g), Kc6.90 x10 If the concentration of H2 in the equilibrium mixture is 0.426 M, what mass of methanol was initially introduced into the vessel? 147 g 74.3 g 33.9 g 49.0 g O 24.8 g

Explanation / Answer

Let the initial conc. of ethanol be a and change in conc. at equilibrium be x .

Now , reaction :

CH3OH(g) <-----> CO(g) + 2H2(g)

Initial a 0 0

Change -x +x +2x

equilibrium a-x x 2x

As kc = [CO] [H2]2/[CH3OH] = x×(2x)2/a-x = 4x3/a-x

Now , equilibrium conc. of H2 = 0.426 M = 2x

So , x = 0.426/2 = 0.213

Therefore , 6.9 ×10-2 = 4×(0.213)3/a-0.213

So , a-0.213 = 0.5602

a = 0.5602 + 0.213 = 0.7732 M = initial conc. of methanol

Now , molar mass of methanol = 32.04 g/mol

Volume of solution = 3.00 L

Therefore , initial mass of methanol used = initial Concentration × Volume of solution × molar mass of methanol

= 0.7732×3×32.04 = 74.3 g .

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