Question

When ethane, $ {C_2}{H_6} $ , reacts with chlorine gas the main product is $ {C_2}{H_5}Cl $ but small amounts of $ {C_2}{H_4}C{l_2} $ are produced. What is the percent yield of $ {C_2}{H_5}Cl $ if the reaction of $ 125g $ of ethane with $ 255g $ of chlorine gas produced $ 206g $ of $ {C_2}{H_5}Cl $ $ ? $

Answer 1

Hint: First we have obtained the balanced chemical reaction. Then using Avogadro’s law obtained the ratio between their reactants. Then convert the reactant masses from grams to moles by using their respective molar masses. Then find the moles of $ {C_2}{H_5}Cl $ and convert them into grams. Finally divide the actual yield by the theoretical yield of chloroethane in grams. Then multiply the difference with $ 100 $ we get the percent yield of $ {C_2}{H_5}Cl $ .

Complete Step By Step Answer:
When ethane ( $ {C_2}{H_6} $ ) reacts with chlorine gas the main product is $ {C_2}{H_5}Cl $ but small amounts of $ {C_2}{H_4}C{l_2} $ are produced, so ignore $ {C_2}{H_4}C{l_2} $ .
The balanced chemical reaction for the chlorination of ethane is written as follows
 $ {C_2}{H_6}\left( g \right) + C{l_2}\left( g \right) \to {C_2}{H_5}Cl\left( g \right) + HCl\left( g \right) $
According to Avogadro’s, one mole of ethane is equal to 1 mole of chlorine gas. Since, the reaction consumes equal numbers of moles of each reactant, which means that chlorine gas will act as a limiting reagent here.
The formula to find the number of moles ( $ n $ ) of a substance of a specific mass ( $ m $ ) in grams is
 $ n = \dfrac{m}{M} $ ----(1)
where, $ M $ is the molar mass of a substance.
Given, $ 125g $ of ethane missing with $ 255g $ of chlorine gas produced $ 206g $ of $ {C_2}{H_5}Cl $ .
Since, we know that the molar mass of $ {C_2}{H_6} = 30.07g $ in one mole $ {C_2}{H_6} $ , the molar mass of $ C{l_2} = 70.91g $ in one mole $ C{l_2} $ , and the molar mass of $ {C_2}{H_5}Cl = 64.51g/mol $ .
Then using the equation (1), we get
 $ \dfrac{{125\;g}}{{30.07g/mol}} = 4.157\;moles\;{C_2}{H_6} $
 $ \dfrac{{255\;g}}{{70.91g/mol}} = 3.596\;moles\;C{l_2} $
Thus, out of the $ 4.157 $ moles of $ {C_2}{H_6} $ , only $ 3.596 $ will actually take part in the reaction.
Because we have a $ 1:1 $ mole ratio between the reactants and chloroethane. This means that the reaction will produce $ 3.596 $ moles of chloroethane.
Use the molar mass of $ {C_2}{H_5}Cl $ to find out how many grams would contain in $ 3.596 $ moles of $ {C_2}{H_5}Cl $ .
Using the equation (1), we get
 $ 3.596 \times 64.51 = 231.98g $ (theoretically)
But the reaction only produced $ 206g $ of $ {C_2}{H_5}Cl $ .
 $ \% \,yield\,of\,{C_2}{H_5}Cl =
\dfrac{{actual\,yield\,of\,{C_2}{H_5}Cl}}{{theoretical\,yield\,of\,{C_2}{H_5}Cl}} \times 100 $
 $ \Rightarrow \% \,yield\,of\,{C_2}{H_5}Cl = \dfrac{{206g}}{{231.98g}} \times 100 = 88.8\% $

Note:
To balance the chemical reaction, consider the unbalanced reaction. Then Identify and count the atoms in both the reactants and products in the following reaction. Then to make the number of atoms the same on both the reactants and products sides, we need to add coefficients (the numbers in front of the formulas) so that the number of atoms of each element is the same on both sides of the chemical equation. Also note that in a double displacement reaction, the chemical bonds between the reactants may be either covalent or ionic.