# What mass of pcl5 will be produced from the given masses of both reactants?

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Need help for Part D

What mass of PCl5 will be produced from the given masses of both
reactants?

Express your answer to three significant figures and include the
appropriate units.   Need help for Part D

What mass of PCl5 will be produced from the given masses of both
reactants?

Express your answer to three significant figures and include the
appropriate units.

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Moles of P4 = mass/molar mass

= 29.0/123.895 = 0.234 mole

Moles of Cl2 = 57.0/71 = 0.802817 mole

Now, balanced equation is–

P4(s) + 10Cl2(g) –> 4PCl5(g)

Moles of P4 required to react completely with Cl2 = (1/10)*moles
of Cl2

= (1/10)*0.802817 = 0.0802817 mole

But P4 present = 0.234 mole ( excess)

Limiting reagent is Cl2 then.

Moles of PCl5 produced = (4/10)*moles of Cl2

= (4/10)*0.802817 = 0.32112 mole

Mass of PCl5 = 0.32112*208.24 = 66.87 gram

= 66.9 gram

If you have any query please comment

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