In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.54 L flask at 1,078 K. At equilibrium, the flask contains 0.113 mol of CO gas, 0.115 mol of H2 gas, and 0.18 mol of methane. What is the water concentration at equilibrium (Kc

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In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.54 L flask at 1,078 K. At equilibrium, the flask contains 0.113 mol of CO gas, 0.115 mol of H2 gas, and 0.18 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1,078)

The water concentration at equilibrium is 0.0202 M

Explanation:

Given the data in the question;

Equilibrium equation; CH₄ + H₂0  ⇄  CO + 3H₂

Equilibrium concentration of CO  [ CO ] = 0.113 / 0.54 = 0.209 M

Equilibrium concentration of H₂    [  H₂ ] = 0.115 / 0.54 = 0.213 M

Equilibrium concentration of CH₄  [ CH₄ ] = 0.18 / 0.54 = 0.333 M

Now, equilibrium constant Kc is;

Kc = [ CO ] [  H₂ ]³ / [ CH₄ ] [ H₂0 ]

we substitute

0.30 = [ 0.209] [ 0.213 ]³ / [ 0.333 ] [ H₂0 ]

0.30 = 0.0020196 / 0.333[ H₂0 ]

0.0999[ H₂0 ] = 0.0020196

[ H₂0 ] = 0.0020196 / 0.0999  

[ H₂0 ] = 0.0202 M

Therefore, the water concentration at equilibrium is 0.0202 M

i would say b

explanation:

but i think i would need more information than that