Question

Through multiple reactions, a chemist uses C, CaO, HCl, and $ {H_2}O $ to produce dichlorobenzene, $ {C_6}{H_4}C{l_2} $ . Assuming that the efficiency of this lab is a minimum of 65.0%, how many molecules of dichlorobenzene could be produced in the lab from 0.500 kg of C?

Answer 1

Hint: This problem can be easily solved by using stoichiometric calculations. We are given that dichlorobenzene is produced from carbon, CaO, HCl and water. By determining the no. of moles of carbon required to produce chlorobenzene we can calculate the no. of molecules of DCB.

Complete Step By Step Answer:
First, we’ll have to find the no. of moles of carbon given to us. Consider the molecular weight of carbon to be 12 g/mol. The reaction given to us has an efficiency of 65.0%. The initial amount of carbon taken is 0.500 kg.
Keeping the rate of efficiency in mind, the amount of carbon present initially will be $ = \dfrac{{\% efficiency}}{{100}} \times mass(g) $
Mass of carbon taken initially will be $ = \dfrac{{65.0}}{{100}} \times (0.500 \times 1000)g = 0.65 \times 500 = 325g $
Now, that we know the initial amount of carbon taken which is 325 grams, we need to find the no. of moles of carbon present initially in the reaction mixture.
 $ Moles = \dfrac{{Mass(g)}}{{M.M}} = \dfrac{{325}}{{12}} = 27mol $
The formation of Chlorobenzene from carbon and chlorine can be given as: $ 6C + 2{H_2} + C{l_2} \to {C_6}{H_4}C{l_2} $
Hence, 6 moles of carbon is needed to form 1 mole of dichlorobenzene. Therefore, 27 moles of carbon will give us: $ \dfrac{{27 \times 1}}{6} = 4.50mol $ of Dichlorobenzene.
Now, we know that one mole of every compound contains $ 6.022 \times {10^{23}} $ atoms or molecules. Therefore, no. of molecules 4.50 moles will have $ = 4.50 \times 6.022 \times {10^{23}} = 2.71 \times {10^{24}} $ molecules of Dichlorobenzene.
Therefore, the no. of molecules of dichlorobenzene that could be produced in the lab from 0.500 kg of Carbon will be $ 2.71 \times {10^{24}} $ molecules. This is the required answer.

Note:
We need just simple stoichiometric calculation to solve this problem. Using the balanced equation, we can find the no. of moles of reactant required to form the desired product. The no. of molecules / atoms in 1 mole of any substance is given by the Avogadro’s number denoted by $ {N_A} $ which has a value of $ 6.022 \times {10^{23}} $ .