This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here's how to calculate the mass of CO(g) at equilibrium: Step 1: Write down the balanced chemical equation and initial moles. The balanced equation is: CO(g) + H_2O(g) CO_2(g) + H_2(g) Initial moles: n(CO) = 0.6 mol n(H_2O) = 0.6 mol n(CO_2) = 0.1 mol n(H_2) = 0.1 mol Volume of flask, V = 2 dm^3 Equilibrium constant, K_c = 4 Step 2: Set up an ICE (Initial, Change, Equilibrium) table for moles. Let x be the change in moles for the reaction. |l|c|c|c|c| & CO(g) & H_2O(g) & CO_2(g) & H_2(g) \\ Initial (mol) & 0.6 & 0.6 & 0.1 & 0.1 \\ Change (mol) & -x & -x & +x & +x \\ Equilibrium (mol) & 0.6-x & 0.6-x & 0.1+x & 0.1+x \\ Step 3: Write the expression for the equilibrium constant, K_c. K_c = [CO_2][H_2][CO][H_2O] Step 4: Substitute equilibrium concentrations into the K_c expression and solve for x. The equilibrium concentrations are calculated by dividing the equilibrium moles by the volume (2 dm^3). K_c = ((0.1+x)/(2))((0.1+x)/(2))((0.6-x)/(2))((0.6-x)/(2)) 4 = ((0.1+x)^2)/(4)((0.6-x)^2)/(4) 4 = ((0.1+x)^2)/((0.6-x)^2) Take the square root of both sides: sqrt(4) = sqrt(((0.1+x)^2)/((0.6-x)^2)) 2 = (0.1+x)/(0.6-x) Now, solve for x: 2(0.6-x) = 0.1+x 1.2 - 2x = 0.1 + x 1.2 - 0.1 = x + 2x 1.1 = 3x x = (1.1)/(3) mol x ≈ 0.3667 mol Step 5: Calculate the moles of CO(g) at equilibrium. n(CO)_eq = 0.6 - x n(CO)_eq = 0.6 - (1.1)/(3) n(CO)_eq = (1.8)/(3) - (1.1)/(3) n(CO)_eq = (0.7)/(3) mol n(CO)_eq ≈ 0.2333 mol Step 6: Calculate the molar mass of CO. Using atomic masses: C = 12.01 g/mol, O = 16.00 g/mol M(CO) = 12.01 + 16.00 = 28.01 g/mol Step 7: Calculate the mass of CO(g) at equilibrium. m(CO) = n(CO)_eq × M(CO) m(CO) = (0.7)/(3) mol × 28.01 g/mol m(CO) = (19.607)/(3) g m(CO) ≈ 6.5356 g Rounding to two decimal places: m(CO) = 6.54 g 3 done, 2 left today. You're making progress.