Question

A gas mixture contains 9.64 mol of C2H4 and 6.66 mol of H2. (a) Compute the mole fraction of C2H4 in the mixture. XC2H4 = (b) A catalyst is then added to the mixture, and the C2H4 starts to react with the H2 to give C2H6: C2H4(g) + H2(g) --> C2H6(g) At a certain point in the reaction, 1.50 mol of C2H6 is present. Determine the mole fraction of C2H4 in the new mixture. XC2H4 =

Answer 1

(a) moles C₂H₄ = 9.64 mol

moles H₂ = 6.66 mol

Total moles = (moles C₂H₄) + (moles H₂)

Total moles = (9.64 mol) + (6.66 mol)

Total moles = 16.30 mol

mole fraction C₂H₄, X_C2H4 = (moles C₂H₄) / (Total moles)

X_C2H4 = (9.64 mol) / (16.30 mol)

X_C2H4 = 0.591

(b) moles C₂H₆ formed = 1.50 mol

moles C₂H₄ consumed = moles C₂H₆ formed

moles C₂H₄ consumed = 1.50 mol

moles C₂H₄ left = (initial moles C₂H₄) - (moles C₂H₄ consumed)

moles C₂H₄ left = (9.64 mol) - (1.50 mol)

moles C₂H₄ left = 8.14 mol

moles H₂ consumed = moles C₂H₆ formed

moles H₂ consumed = 1.50 mol

moles H₂ left = (initial moles H₂) - (moles H₂ consumed)

moles H₂ left = (6.66 mol) - (1.50 mol)

moles H₂ left = 5.16 mol

Total moles = (moles C₂H₆ formed) + (moles C₂H₄ left) + (moles H₂ left)

Total moles = (1.50 mol) + (8.14 mol) + (5.16 mol)

Total moles = 14.80 mol

mole fraction C₂H₄, X_C2H4 = (moles C₂H₄ left) / (Total moles)

X_C2H4 = (8.14 mol) / (14.80 mol)

X_C2H4 = 0.550