Question

The formation of polyethene from calcium carbide takes place as follows:
$Ca{C_2} + 2{H_2}O \to Ca{\left( {OH} \right)_2} + {C_2}{H_2}$
${C_2}{H_2} + {H_2} \to {C_2}{H_4}$
$n{C_2}{H_4} \to {\left( {C{H_2} - C{H_2}} \right)_n}$
The amount of polyethene obtained from $64$ kg of $Ca{C_2}$ is?
A) $14$ kg
B) $7$ kg
C) $21$ kg
D) $28$ kg

Answer 1

From the given equations, it is clear that the product formed is a polymer. Balance the given three equations and find the amount of ethyne produced for one mole of $Ca{C_2}$ which helps us find the amount of ethyne produced when $64$ kg of $Ca{C_2}$.

Complete step-by-step solution: Given to us are three equations. In the first equation, $Ca{C_2}$ is reacting with water to form ethyne. The balanced equation for this reaction is written as $Ca{C_2} + 2{H_2}O \to Ca{\left( {OH} \right)_2} + {C_2}{H_2}$
It is given to us that $64$ kg of $Ca{C_2}$ is reacting to form ethyne. Now, from the above reaction it is clear that one mole of $Ca{C_2}$ is producing one mole of ethyne.This means that $64$ grams of $Ca{C_2}$ produces $26$ grams of ethyne. Let us assume that $64$ kg or $64 \times 1000$ grams of $Ca{C_2}$ produces X grams of ethyne. The value of X would be $X = \dfrac{{64 \times 1000 \times 26}}{{64}} = 26 \times 1000$ grams. Hence $26$ kgs of ethyne is produced. Now the second reaction is the formation of ethene from ethyne. This reaction is given as ${C_2}{H_2} + {H_2} \to {C_2}{H_4}$
Now we have $26$ kgs of ethyne. One mole of ethyne produces one mole of ethene and this means that $26$ grams of ethyne produces $28$ grams of ethene.From this we can conclude that $26$ kgs of ethyne will produce $28$ kgs of ethene.The final reaction is the polymerization of ethene. We can write this reaction as follows.
$n{C_2}{H_4} \to {\left( {C{H_2} - C{H_2}} \right)_n}$
From this reaction we can conclude that $28$ kgs of ethene would produce $28$ kgs of polythene.

Therefore, the correct answer is option D).

Note: It is to be noted that the equations of the reactions should be balanced before taking the stoichiometric coefficients from them. In the above reactions the ratio of number of moles of reactant and product are $1:1$