Question

If 6.0 g of fe(co)5 reacts with 4.0 g of pf3 and 4.0 g of h2, which is the limiting reagent

Answer 1

Number of moles of Fe(CO)5 = mass/molar mass = (6.0 g) / (195.8955 g/mol) = 0.03 mol Number of moles of PF3 = (4.0 g)/ (87.97 g/mol) = 0.0455 mol Number of moles of H2 = (4.0 g) / (2.016 g/mol) = 1.984 mol From the balanced equation 1 mol of Fe(CO)5 reacts with 2 mol of PF3 and 1 mol of H2 then 0.03 mol of Fe(CO)5 will need 0.03 (2/1) = 0.06 mol of PF3 and 0.03 mol of H2 Since we have less PF3 ( 0.0455 mol instead of 0.06 mol), that is the limiting agent