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Suppose a 500. mL flask is filled with 0.10 mol of CO, 1.3 mol of H₂O and 1.5 mol of H₂. This reaction becomes possible: CO(g) + H₂O(g)CO₂(g) + H₂(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of CO₂. You can leave out the M symbol for molarity. initial change equilibrium CO 0 0 H₂O CO₂ 0 x H₂ 0 0 0

Suppose a 500. mL flask is filled with 0.10 mol of CO, 1.3 mol of H₂O and 1.5 mol of H₂. This reaction becomes possible: CO(g) + H₂O(g)CO₂(g) + H₂(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of CO₂. You can leave out the M symbol for molarity. initial change equilibrium CO 0 0 H₂O CO₂ 0 x H₂ 0 0 0

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Suppose a 500. mL flask is filled with 0.10 mol of CO, 1.3 mol of H₂O and 1.5 mol of H₂. This reaction becomes possible: CO(g) + H₂O(g) → CO₂(g) + H₂(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of CO2. You can leave out the M symbol for molarity. initial change equilibrium CO 0 H₂O 0 CO2 0 X 0 H₂ 0 0/0 X Ś
Transcribed Image Text:Suppose a 500. mL flask is filled with 0.10 mol of CO, 1.3 mol of H₂O and 1.5 mol of H₂. This reaction becomes possible: CO(g) + H₂O(g) → CO₂(g) + H₂(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of CO2. You can leave out the M symbol for molarity. initial change equilibrium CO 0 H₂O 0 CO2 0 X 0 H₂ 0 0/0 X Ś
Expert Solution
Step 1

The equilibrium reaction is as follows:

CO(g) + H2O(g) ⇌ CO2(g) + H2(g) 

The moles of CO = 0.10 mol

The moles of H2O = 1.3 mol

The moles of H2 = 1.5 mol

 

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Try Again Your answer is incorrect. • CO: Your initial concentration value is incorrect. H₂ O: Your initial concentration value is incorrect. CO₂: Your initial concentration value is incorrect. H₂: Your initial and equilibrium concentration values are incorrect. ● ● Suppose a 500. mL flask is filled with 0.10 mol of CO, 1.3 mol of H₂O and 1.5 mol of H₂. This reaction becomes possible: CO(g) + H₂O(g) → CO₂(g) + H₂(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of CO₂. You can leave out the M symbol for molarity. initial H₂O 0.2 M 2.6M 38888 2.6 - x change equilibrium CO -X CO₂ 0.2 - X H₂ X 3M +x X 00 X S
Transcribed Image Text:Try Again Your answer is incorrect. • CO: Your initial concentration value is incorrect. H₂ O: Your initial concentration value is incorrect. CO₂: Your initial concentration value is incorrect. H₂: Your initial and equilibrium concentration values are incorrect. ● ● Suppose a 500. mL flask is filled with 0.10 mol of CO, 1.3 mol of H₂O and 1.5 mol of H₂. This reaction becomes possible: CO(g) + H₂O(g) → CO₂(g) + H₂(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of CO₂. You can leave out the M symbol for molarity. initial H₂O 0.2 M 2.6M 38888 2.6 - x change equilibrium CO -X CO₂ 0.2 - X H₂ X 3M +x X 00 X S
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