Question

Balance the following reaction.
$Fe{C_2}{O_4} + KMn{O_4} + {H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + MnS{O_4} + {K_2}S{O_4} + {H_2}O$
A. $10Fe{C_2}{O_4} + 6KMn{O_4} + 24{H_2}S{O_4} \to 5F{e_2}{\left( {S{O_4}} \right)_3} + 20C{O_2} + 6MnS{O_4} + 3{K_2}S{O_4} + 24{H_2}O$
B. $5Fe{C_2}{O_4} + 4KMn{O_4} + 12{H_2}S{O_4} \to 7F{e_2}{\left( {S{O_4}} \right)_3} + 20C{O_2} + 6MnS{O_4} + 3{K_2}S{O_4} + 24{H_2}O$
C. $10Fe{C_2}{O_4} + 4KMn{O_4} + 24{H_2}S{O_4} \to 6F{e_2}{\left( {S{O_4}} \right)_3} + 20C{O_2} + 6MnS{O_4} + 3{K_2}S{O_4} + 24{H_2}O$
D. None of these

Answer 1

Hint: We know that chemical equations in which the number of atoms of all the involved species is equal on reactant as well as the product side are called balanced chemical equations.

Complete step by step answer:
We use chemical equations to represent chemical reactions by using the chemical formulae of all the involved species. For example, burning magnesium ribbon to give magnesium oxide can simply be represented with the following chemical equation:
$Mg + {O_2} \to MgO$
We know that mass is conserved during any chemical reaction which can also be stated using Dalton’s theory as atoms can neither be created nor destroyed but only reorganized in a chemical reaction. From this, it can be inferred that the total number of atoms of each element must remain the same giving rise to balanced chemical equations.

Now, let’s try to balance the given reaction by trial and error method as follows:
The given reaction is:
$Fe{C_2}{O_4} + KMn{O_4} + {H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + MnS{O_4} + {K_2}S{O_4} + {H_2}O$

Let’s start by calculating the number of atoms for each element on both sides:
$Fe{C_2}{O_4} + KMn{O_4} + {H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + MnS{O_4} + {K_2}S{O_4} + {H_2}O$

	Fe	C	O	K	Mn	H	S
Reactant side	1	2	12	1	1	2	1
Product side	2	1	23	2	1	2	5

Let’s start with sulfur atoms that are five on the product side but only one at the reactant side. So, we will multiply ${H_2}S{O_4}$ on the reactant side by five as follows:
$Fe{C_2}{O_4} + KMn{O_4} + 5{H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + MnS{O_4} + {K_2}S{O_4} + {H_2}O$

	Fe	C	O	K	Mn	H	S
Reactant side	1	2	28	1	1	10	5
Product side	2	1	23	2	1	2	5

Now, for hydrogen atoms that are ten on the reactant side but only two at the product side. So, we will multiply ${H_2}O$ on the product side by five as follows:
$Fe{C_2}{O_4} + KMn{O_4} + 5{H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + MnS{O_4} + {K_2}S{O_4} + 5{H_2}O$

	Fe	C	O	K	Mn	H	S
Reactant side	1	2	28	1	1	10	5
Product side	2	1	27	2	1	10	5

Now, for potassium atoms that are two on the product side but only one at the reactant side. So, we will multiply the reactant side by two as follows:
\[Fe{C_2}{O_4} + 2KMn{O_4} + 5{H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + MnS{O_4} + {K_2}S{O_4} + 5{H_2}O\]

	Fe	C	O	K	Mn	H	S
Reactant side	1	2	32	2	2	10	5
Product side	2	1	27	2	1	10	5

Now, for manganese atoms that are two on the reactant side but only one at the product side. So, we will multiply \[MnS{O_4}\] on the product side by two as follows:
$Fe{C_2}{O_4} + 2KMn{O_4} + 5{H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + 2MnS{O_4} + {K_2}S{O_4} + 5{H_2}O$

	Fe	C	O	K	Mn	H	S
Reactant side	1	2	32	2	2	10	5
Product side	2	1	31	2	2	10	6

Now, for iron atoms that are two on the product side but only one at the reactant side. So, we will multiply $Fe{C_2}{O_4}$ on the reactant side by two as follows:
$2Fe{C_2}{O_4} + 2KMn{O_4} + 5{H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + 2MnS{O_4} + {K_2}S{O_4} + 5{H_2}O$

	Fe	C	O	K	Mn	H	S
Reactant side	2	4	36	2	2	10	5
Product side	2	1	31	2	2	10	6

Again, for sulfur atoms that are six on the product side but only five at the reactant side. So, making them six on the reactant side as well:
$2Fe{C_2}{O_4} + 2KMn{O_4} + 6{H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + 2MnS{O_4} + {K_2}S{O_4} + 5{H_2}O$

	Fe	C	O	K	Mn	H	S
Reactant side	2	4	40	2	2	12	6
Product side	2	1	31	2	2	10	6

Again, for hydrogen atoms that are twelve on the reactant side but only ten at the product side. So, making them twelve on the product side as well:
$2Fe{C_2}{O_4} + 2KMn{O_4} + 6{H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + C{O_2} + 2MnS{O_4} + {K_2}S{O_4} + 6{H_2}O$

	Fe	C	O	K	Mn	H	S
Reactant side	2	4	40	2	2	12	6
Product side	2	1	32	2	2	12	6

Now, for carbon atoms that are four on the reactant side but only one at the product side. So, making them four on the product side as well:
\[2Fe{C_2}{O_4} + 2KMn{O_4} + 6{H_2}S{O_4} \to F{e_2}{\left( {S{O_4}} \right)_3} + 4C{O_2} + 2MnS{O_4} + {K_2}S{O_4} + 6{H_2}O\]

	Fe	C	O	K	Mn	H	S
Reactant side	2	4	40	2	2	12	6
Product side	2	4	38	2	2	12	6

Now, for oxygen atoms that are forty on the reactant side but only thirty eight at the product side. So, balancing them on the both sides:
\[10Fe{C_2}{O_4} + 6KMn{O_4} + 24{H_2}S{O_4} \to 5F{e_2}{\left( {S{O_4}} \right)_3} + 20C{O_2} + 6MnS{O_4} + 3{K_2}S{O_4} + 24{H_2}O\]

	Fe	C	O	K	Mn	H	S
Reactant side	10	20	160	6	6	48	24
Product side	10	20	160	6	6	48	24

Hence, the balanced equation is \[10Fe{C_2}{O_4} + 6KMn{O_4} + 24{H_2}S{O_4} \to 5F{e_2}{\left( {S{O_4}} \right)_3} + 20C{O_2} + 6MnS{O_4} + 3{K_2}S{O_4} + 24{H_2}O\] and the correct option is A.

Therefore, from the above explanation the correct option is (A).

Note:Skeletal chemical equations do not follow the law of conservation of mass and can give information about formulae only but in balanced chemical equations, law of conservation is followed and we can get stoichiometric molar relationships as well. We have to be careful while calculating and balancing the number of atoms.