9) A 12.39 kg sample of phosphorus reacts with 52.54 kg of chlorine to form only

Question

9) A 12.39 kg sample of phosphorus reacts with 52.54 kg of chlorine to form only phosphorus trichloride (PCl3).

a) If it is the only product, what mass of PCl3 is formed?

b) Assuming ideal behavior, calculate the stoichiometric volume of chlorine gas needed at 1.1 atm and 25 °C. 4

c) Calculate the stoichiometric mass of chlorine required.

d) In a similar process, the chlorine is introduced to the reactor in the form of a carbon tetrachloride solution. What volume of a 4.5 M solution is required to deliver for the stoichiometric amount of chlorine?

Explanation / Answer

the reaction is

P + 1.5 Cl2 ---> PCl3

now

moles = mass / molar mass

so

moles of P = 12390 / 31

moles of P = 399.677

now

moles of Chlorine = 52540 / 71

moles of Chlorine = 740

now

moles of Chlorine required = 1.5 x moles of P

moles of chlorine required = 1.5 x 399.677

moles of chlorine required = 600

but

740 moles of chlorine

so

chlorine is in excess

a)

now

moles of PCl3 = moles of P reacted = 400

mass = moles x molar mass

so

mass of PCl3 = 400 x 137.33

mass of PCl3 formed = 54.93 kg

b)

PV = nRT

1.1 x V = 600 x 0.0821 x 298

V = 13345

so

13345 L of chlorine gas is required

c)

mass = moles x molar mass

so

mass of Chlorine required = 600 x 71

mass of chlortine required = 42.6 kg

d)

now

moles of CCl4 required = 2 x moles of CL2

moles of CCl4 required = 2 x 600

moles of CCL4 required = 1200

now

volume = moles / molarity

so

volume (L ) = 1200 / 4.5

volume (L) = 266.666

so

266.66 L of solution is required

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