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How many moles of #PCL_5# can be produced from 22.0 g of #P_4# (and excess #Cl_2)#?

Answer 1

Noah Aldrich

Approx. #0.7*mol# with respect to #PCl_5#.

To depict the reaction, we require (i) a stoichiometric equation:

#1/4P_4(s) + 5/2Cl_2(g) rarr PCl_5(s)#

Integral numbers can be obtained by multiplying the coefficients out if desired:

#P_4(s) + 10Cl_2(g) rarr 4PCl_5(s)#

If you use the earlier equation, I think handling the multiplication will be a little simpler.

Additionally, we require (ii) equal amounts of every reactant:

Moles of #P_4# #=# #(22.0*g)/(4xx31.00*g*mol^-1)=0.177*mol#

There is an abundance of the gas oxidant chlorine.

And thus, by the given stoichiometry with respect to phosphorus, we can make #0.177*molxx4=0.710*mol# with respect to #PCl_5#.

This represents a mass of #0.177*molxx4xx208.24*g*mol^-1=??g#

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Answer 2

Caleb Adams

From 22.0 g of P3 (and excess Cl3₂), 5.5 moles of PCl₅ can be made.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.