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Question: Given \(2C{{l}_{2}}(g)+{{C}_{2}}{{H}_{2}}(g)\to {{C}_{2}}{{H}_{2}}C{{l}_{4}}(l)\). How many litres o...

Given 2Cl2(g)+C2H2(g)→C2H2Cl4(l)2C{{l}_{2}}(g)+{{C}_{2}}{{H}_{2}}(g)\to {{C}_{2}}{{H}_{2}}C{{l}_{4}}(l). How many litres of chlorine will be needed to make 75.0 grams of C2H2Cl4{{C}_{2}}{{H}_{2}}C{{l}_{4}} ? Pressure: 1atm, Temperature: 298K.

Explanation

Solution

Based on the data given, find out the mole ratio between the two compounds and by using the mass of tetrachloromethane determine how many moles are being produced and finally deduce the answer using the ideal gas equation.

Complete step by step solution:
In the lower classes of physical chemistry, we have studied the basic topics such as about the Vander Waals equation and also regarding the basic ideal gas equations and some of the related derivations and equations.
Let us make use of a few such concepts and approaches for the required answer.
Here, we have the balanced chemical equation which is as follows,
2Cl2(g)+C2H2(g)→C2H2Cl4(l)2C{{l}_{2}}(g)+{{C}_{2}}{{H}_{2}}(g)\to {{C}_{2}}{{H}_{2}}C{{l}_{4}}(l)
Now, let us see the mole ratio between these two reaction sides.
- Here, from the balanced equation given above, we can conclude that 2 moles of chlorine gas reacts with 1 mole of ethane to produce one mole tetrachloroethane. Thus, the mole ratio of this can be said as, 2 :1.
- Now, by knowing the mass of tetrachloromethane, we shall calculate the number of moles produced and that will be, 75×1168=0.44675\times \dfrac{1}{168}=0.446 moles of C2H2Cl4{{C}_{2}}{{H}_{2}}C{{l}_{4}}.
Based on the mole ratio let us calculate total moles of chlorine required to produce the necessary product and this will be,
0.446moles(C2H2Cl4)×2moles Cl21mole(C2H2Cl4)=0.892moles(Cl2)0.446moles({{C}_{2}}{{H}_{2}}C{{l}_{4}})\times \dfrac{2moles\, Cl_2}{1mole({{C}_{2}}{{H}_{2}}C{{l}_{4}})}=0.892moles(C{{l}_{2}})
The volume of the reactant that is chlorine required can be calculated using the ideal gas equation which is given by, PV=nRTPV=nRT
This, equation can be rearranged as, V=nRTPV=\dfrac{nRT}{P}
Here, n is the number of moles. Substituting the values in above equation, we get
V=0.892×0.082(L.atmmol−K)×298K1atm=21.8LV=\dfrac{0.892\times 0.082\left( \dfrac{L.atm}{mol-K} \right)\times 298K}{1atm}=21.8L
Thus, 21.8L of chlorine required to make 75g of C2H2Cl4{{C}_{2}}{{H}_{2}}C{{l}_{4}}

Note:
Note that ideal gas equation which is also called general gas equation is applicable for many gases which determine their behaviour under many conditions and this is the equation of a state of hypothetical ideal gas.