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

1 answer:

5 0

<span>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</span>

You might be interested in

Which statement best describes the effect of low ionization energies and low electronegativities on metallic bonding?

kramer

The statement that best describes the effect of low ionization energies and low electronegativities on metallic bonding is the first one - the valence electrons are easily delocalized.
Due to these low energies and negativities, valence electrons can be moved around quite easily and their positions may be altered quite drastically.

6 0

3 years ago

Read 2 more answers

Net Ionic equations

Goryan [66]

<span>Net Ionic equation: 3Zn^2+(aq) + 2PO4^3-(aq) ---> Zn3(PO4)2 (s)
hope it helps
</span>

4 0

3 years ago

How many significant figures in 11 soccer players

maria [59]

Counting gives an exact number and exact numbers have infinite sig figs.

3 0