Question

from the following reaction
$2Co{F_2} + {F_2} \to 2Co{F_3}$
${\left( {C{H_2}} \right)_n} + 4nCo{F_3} \to {\left( {C{F_2}} \right)_n} + 2nHF + 4nCo{F_2}$
Calculate how much ${F_2}$ will consumed to produced $1kg$ of ${\left( {C{F_2}} \right)_n}$
$\left( A \right)1.52kg$
$\left( B \right)2.04kg$
$\left( C \right)0.76kg$
$\left( D \right)4.56kg$

Answer 1

Hint: Firstly check if both the equations are balanced or not . After balancing the equation we have seen how many moles are used while forming the products. As we need ${\left( {C{F_2}} \right)_n}$ so we need to make a relation where we can check the consumption of fluorine while forming this product ,because we have provide $F = 19$

Complete step by step answer:
$\left( 1 \right)$ $2Co{F_2} + {F_2} \to 2Co{F_3}$
$ \Rightarrow $ From the above equation 1 mole of ${F_2} \equiv $2 mole of $Co{F_3}$
$\left( 2 \right)$ $2n \times \dfrac{{1000g}}{{50n}} = 40mole$${\left( {C{H_2}} \right)_n} + 4nCo{F_3} \to {\left( {C{F_2}} \right)_n} + 2nHF + 4nCo{F_2}$
$ \Rightarrow $ From above equation 4n mole of $Co{F_3}$$ \equiv $ 1 mole of ${\left( {C{F_2}} \right)_n}$
 Now , 1 mole of ${F_2} \equiv $2 mole of $Co{F_3}$
$ \Rightarrow $ needs to multiply 2n to the relation to make the value of $Co{F_3}$ equal .
$2n{F_2} \equiv 4nCo{F_3}$
 $ \Rightarrow $ we can write $2n{F_2} \equiv {\left( {C{F_2}} \right)_n}$
$1{\left( {C{F_2}} \right)_n} = 50n$
 $2n \times \dfrac{{1000g}}{{50n}} = 40mole$
Weight of ${F_2}$ given $ = 19$
Weight of the ${F_2}$required $40 \times 38 = 1520g$ $ = 1.52kg$

So, option $A$ is correct .
Note:
To balance a chemical equation we need to focus on two things: the subscript and coefficients of the different elements . Subscript is a part of a formula and it can’t be changed whereas coefficient is the number of particular reactants present. Coefficient is regarded as the number of moles , it is present just in front of any chemical formula. If no coefficient is present then it is regarded as 1 mole . mole is a unit of measurement for the amount of substance. so that’s why we wrote 1 mole of fluorine is almost equal to $Co{F_3}$.
Molecular weight of fluorine is 19
Molecular weight of carbon is 12
Molecular weight of cobalt 59
${\left( {C{F_2}} \right)_n}$ so putting the values we get $50n$
In a chemical reaction just like energy , the atoms can neither be created nor destroyed . so even if the reactions are changed the number of the atoms in the reactant and product will always be equal.