
Balance the following reaction and mention the states of reactants and products.
$ KMn{O_4} + HCl \to KCl + MnC{l_2} + {H_2}O + C{l_2} $
Answer
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Hint :We must ensure that the number of atoms of each element on the reactant side equals the number of atoms of each element on the product side in order to balance the chemical equation. We must multiply the number of atoms in each product so all sides are equal in order to make both sides equal.
Complete Step By Step Answer:
Here, the given equation is,
$ KMn{O_4} + HCl \to KCl + MnC{l_2} + {H_2}O + C{l_2} $
Therefore,
$ KMn{O_4} + HCl $ is the reactant side and $ KCl + MnC{l_2} + {H_2}O + C{l_2} $ is the products side. In the reactant side there are 1 K atom, 1 Mn atom , 4 O atom , 1 H atom and 1 Cl atom. But in the reactant side, there are 1 K atom, 1 Mn atom ,1 O atom , 1 H atom and 1+2+ 2= 5 Cl atoms. Hence, the number of atoms is not balanced on both sides.
To balance the atoms on both sides, we have to add coefficients; Coefficients are the numbers that are set in front of formulas to balance quantities, and they multiply all of the atoms in the formula.
First, we must balance the oxygen atoms,
$ KMn{O_4} + HCl \to KCl + MnC{l_2} + 4{H_2}O + C{l_2} $ , the O atom is balanced
Then, we must balance the hydrogen atoms,
$ KMn{O_4} + 8HCl \to KCl + MnC{l_2} + 4{H_2}O + C{l_2} $ , the H atom is balanced
Now, we can balance Cl and H atoms,
$ KMn{O_4} + 8HCl \to KCl + MnC{l_2} + 4{H_2}O + \dfrac{5}{2}C{l_2} $
Now, multiply by 2 on both the sides of the equation,
$ 2(KMn{O_4} + 8HCl) \to 2(KCl + MnC{l_2} + 4{H_2}O + \dfrac{5}{2}C{l_2}) $
$ 2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2} $
That is,
Now we have 2 K atom, 2 Mn atom , 8 O atom , 16 H atom and 16 Cl atom on both the sides.
Hence, the equation is balanced.
Therefore, the balanced final equation is ,
$ 2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2} $
Note :
When the reactant and product sides of an equation have the same number of each element, the equation is balanced. To correctly represent the law of conservation of matter, equations must be balanced.
Complete Step By Step Answer:
Here, the given equation is,
$ KMn{O_4} + HCl \to KCl + MnC{l_2} + {H_2}O + C{l_2} $
Therefore,
$ KMn{O_4} + HCl $ is the reactant side and $ KCl + MnC{l_2} + {H_2}O + C{l_2} $ is the products side. In the reactant side there are 1 K atom, 1 Mn atom , 4 O atom , 1 H atom and 1 Cl atom. But in the reactant side, there are 1 K atom, 1 Mn atom ,1 O atom , 1 H atom and 1+2+ 2= 5 Cl atoms. Hence, the number of atoms is not balanced on both sides.
To balance the atoms on both sides, we have to add coefficients; Coefficients are the numbers that are set in front of formulas to balance quantities, and they multiply all of the atoms in the formula.
First, we must balance the oxygen atoms,
$ KMn{O_4} + HCl \to KCl + MnC{l_2} + 4{H_2}O + C{l_2} $ , the O atom is balanced
Then, we must balance the hydrogen atoms,
$ KMn{O_4} + 8HCl \to KCl + MnC{l_2} + 4{H_2}O + C{l_2} $ , the H atom is balanced
Now, we can balance Cl and H atoms,
$ KMn{O_4} + 8HCl \to KCl + MnC{l_2} + 4{H_2}O + \dfrac{5}{2}C{l_2} $
Now, multiply by 2 on both the sides of the equation,
$ 2(KMn{O_4} + 8HCl) \to 2(KCl + MnC{l_2} + 4{H_2}O + \dfrac{5}{2}C{l_2}) $
$ 2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2} $
That is,
Now we have 2 K atom, 2 Mn atom , 8 O atom , 16 H atom and 16 Cl atom on both the sides.
Hence, the equation is balanced.
Therefore, the balanced final equation is ,
$ 2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2} $
Note :
When the reactant and product sides of an equation have the same number of each element, the equation is balanced. To correctly represent the law of conservation of matter, equations must be balanced.
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