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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

Jeremy Wood

in Chemistry

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Carlton Burgess on May 1, 2018

The number of moles of Fe(CO)5 = mass/molar mass = (6.0 g) / (195.8955 g/mol) = 0.03 mol The number of moles PF3 = (4.0 g)/ (87.97 g/mol) = 0.0455 mol The 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 moles of PF3 and 1 mol of H2 then 0.03 mol of Fe(CO)5 will have 0.03 (2/1) = 0.06 mol of PF3 and 0.03 mol of H2 Since we have less PF3 ( 0.0455 moles instead of 0.06 mol), which is the limiting agent


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