Pentacarbonyliron(0) (fe(co)5) reacts with phosphorous trifluoride (pf3) and hydrogen, releasing carbon monoxide: fe(co)5+2pf3+h2→fe(co)2(pf3)2(h)2+3co 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?

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barnuts barnuts · Answer 1 · 2017-11-22T12:27:51+01:00

The limiting reactant can be determined by calculating the moles supplied / moles stoich ratio and the lowest is the limiting reactant.

Fe(CO)5 ratio = [6 g / 195.9 g/mol] / 1

Fe(CO)5 ratio = 0.0306

PF3 ratio = [4 g / 87.97 g/mol] / 2

PF3 ratio = 0.0227

H2 ratio = [4 g / 2 g/mol] / 1

H2 ratio = 2

We can see that PF3 has the lowest ratio, so it is the limiting reactant.

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Oseni Oseni · Answer 2 · 2022-06-01T15:45:33+02:00

The limiting reagent in the reaction will be [tex]PF_3[/tex]

They are reagents that limit the extent to which reactions can go in terms of the amount of products yielded.

From the equation of the reaction:

[tex]Fe(CO)_5 + 2PF_3 + H_2--- > Fe(CO)_2(PF_3)_2(H_2) + 3CO[/tex]

The mole ratio of [tex]Fe(CO)_5[/tex] , [tex]PF_3[/tex] and [tex]H_2[/tex] is 1:2:1.

Mole of 6.0 g [tex]Fe(CO)_5[/tex] = 6/195.9 = 0.03 moles

Mole of 4.0 g [tex]PF_3[/tex] = 4/87.97 = 0.04

Mole of 4.0 g [tex]H_2[/tex] = 4/2 = 2 moles

Thus, [tex]PF_3[/tex] is the limiting reagent.

More on limiting reagents can be found here:https://brainly.com/question/11848702

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