Question

Formation of polyethylene from calcium carbide takes place as follows:
$\begin{aligned} & Ca{{C}_{2}}+2{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}}+{{C}_{2}}{{H}_{2}} \\\ & {{C}_{2}}{{H}_{2}}+{{H}_{2}}\to {{C}_{2}}{{H}_{4}} \\\ & n{{C}_{2}}{{H}_{4}}\to \left( - \right.C{{H}_{2}}-C{{H}_{2}}{{\left. - \right)}_{n}} \\\ \end{aligned}$
The amount of polyethylene obtained from 64.0 Kg of $Ca{{C}_{2}}$is:
(A) 27 Kg
(B) 24 Kg
(C) 22 Kg
(D) 28 Kg

Answer 1

The end product of the reaction is the polymer. By proper balancing of the equations gives the amount of formation of the polymer.

Complete step by step solution:
Let us first see about the balancing of the equations and then solve the given problem,
Balancing the equation-
The equation will be called balanced when the number of atoms of the reactants will be equal to the number of atoms of the product.
This can be done by putting the stoichiometric coefficients to the reaction on both sides. The subscripts are never changed to balance the equation.
After balancing the equation, we need to solve for the amounts of the reactant or product present in the reaction by including its molar weights and number of moles present in the given criteria.
Thus, now solving for the given illustration we get;
The equations are,
$\begin{aligned} & Ca{{C}_{2}}+2{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}}+{{C}_{2}}{{H}_{2}} \\\ & {{C}_{2}}{{H}_{2}}+{{H}_{2}}\to {{C}_{2}}{{H}_{4}} \\\ & n{{C}_{2}}{{H}_{4}}\to \left( - \right.C{{H}_{2}}-C{{H}_{2}}{{\left. - \right)}_{n}} \\\ \end{aligned}$
We can see that the equations are already balanced, thus now we need to consider the given masses of any of the components to find the amount of polyethene obtained at the end.
Thus, given that the mass of $Ca{{C}_{2}}$ is 64 Kg.
Now,
The molar mass of the $Ca{{C}_{2}}$ is 64 g/mol.
64 Kg of $Ca{{C}_{2}}$ = $\dfrac{64\times 1000g}{64g/mol}$ = 1000 mol.
By stoichiometry of the reaction,
1000 moles of $Ca{{C}_{2}}$ = 1000 moles of ${{C}_{2}}{{H}_{2}}$
1000 moles of ${{C}_{2}}{{H}_{2}}$ = 1000 moles of ${{C}_{2}}{{H}_{4}}$
Now, the molar mass of ${{C}_{2}}{{H}_{4}}$ is 28 g/mol.
Thus,
1000 moles of ${{C}_{2}}{{H}_{4}}$ = 1000 moles $\times$ 28 g/mol = 28 Kg.
Again, observing the last equation where the polymer is formed and including stoichiometry we get;
28 Kg of ${{C}_{2}}{{H}_{4}}$ will give 28 Kg of polyethene.

Hence, option (D) is correct.

Note: Do note that when the equation is not balanced, we need to balance them by including the stoichiometric coefficients and not changing the subscripts of the molecules.