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tiny-mole [99]
3 years ago
13

In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.778 L

flask at 1,072 K. At equilibrium, the flask contains 0.206 mol of CO gas, 0.187 mol of H2 gas, and 0.187 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1,072 K)? Enter to 4 decimal places. HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the Kc expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into Kc expression, solve for the unknown.
Chemistry
1 answer:
sergejj [24]3 years ago
7 0

Answer:

the water concentration at equilibrium is

⇒ [ H2O(g) ] = 0.0510 mol/L

Explanation:

  • CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)

∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30

  • equilibrium:

⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L

⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

replacing in Kc:

⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30

⇒ 0.0721 [ H2O(g) ] = 3.679 E-3

⇒ [ H2O(g) ] = 0.0510 mol/L

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Decomposition reaction -

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Cu + 2 AgNO3 + Cu(NO3)2 + 2Ag
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To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...

Limiting reactant determination:

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