Pentacarbonyliron (Fe(CO)₅) reacts with phosphorous trifluoride (PF₃) and hydrogen, releasing carbon monoxide: The reaction of 5.0 mol of Fe(CO)₅, 8.0 mol of PF₃and 6.0 mol of H₂will release _ _ mol of CO. A) 24 B) 12 C) 6.0 D) 15 E) 5.0

Question

Quizplus · Accepted Answer

According to the balanced equation, one mole of Fe(CO)₅reacts with three moles of PF₃and six moles of H₂to release five moles of CO. Therefore, 5.0 mol of Fe(CO)₅reacts with (3/1) x 8.0 mol = 24.0 mol of PF₃and (6/1) x 6.0 mol = 36.0 mol of H₂. The limiting reagent is Fe(CO)₅, and it will react completely with 3 x 5.0 mol = 15.0mol of PF₃, releasing 3 x 5.0 mol = 15.0 mol of CO. Therefore, the answer is 15.0 mol of CO, which is equivalent to (15.0/5.0) = 3 times the stoichiometric amount of CO. Hence, the correct answer is B.

Pentacarbonyliron (Fe(CO)5) Reacts with Phosphorous Trifluoride (PF3) and Hydrogen, Releasing

Pentacarbonyliron (Fe(CO)₅) Reacts with Phosphorous Trifluoride (PF₃) and Hydrogen, Releasing