C 6 H 6(g) + 3H 2(g) <-------> C 6 H 12(g) When 1.00 moles of C 6 H 6(g) and 3.0

Question

C6H6(g) + 3H2(g)<-------> C6H12(g) When 1.00 moles of C6H6(g) and 3.00moles H2 are put into a 200. L container and allowed toreach equilibrium over a catalyst at elevated temperature, theresulting mixture contains 0.137 mol . What is theequilibrium amount of H2 in moles.? C6H6(g) + 3H2(g)<-------> C6H12(g) When 1.00 moles of C6H6(g) and 3.00moles H2 are put into a 200. L container and allowed toreach equilibrium over a catalyst at elevated temperature, theresulting mixture contains 0.137 mol . What is theequilibrium amount of H2 in moles.?

Explanation / Answer

                                C6H6(g)         +      3H2(g)     <------->     C6H12(g)                 Initial: 1.00mol/2.00L           3.00mol/2.00L              0 mol/L             Change: - 0.137 mol / 2.00 L   - 3(0.137 mol / 2.00L)    + 0.137mol/2.00L          Equilibrium:0.4315mol/L             1.2945mol/L                    0.0685 mol/L      Equilibrium concentration of H2 = 1.2945mol/L The equilibrium amount in moles of H2 = 1.295mol/L * 2.00 L                                                            = 2.59 mol                 Initial: 1.00mol/2.00L           3.00mol/2.00L              0 mol/L             Change: - 0.137 mol / 2.00 L   - 3(0.137 mol / 2.00L)    + 0.137mol/2.00L          Equilibrium:0.4315mol/L             1.2945mol/L                    0.0685 mol/L      Equilibrium concentration of H2 = 1.2945mol/L The equilibrium amount in moles of H2 = 1.295mol/L * 2.00 L                                                            = 2.59 mol

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