
How would you balance the following equation
$NH_{3}+O_{2}\rightarrow NO_{2}+H_{2}O$
Answer
542.4k+ views
Hint: This reaction is the formation of nitrogen dioxide from ammonia. In order to balance the equation, first balance the nitrogen and then balance the hydrogen and oxygen atoms.
Complete answer:
- The reaction given in the question is the formation of nitrogen dioxide from ammonia when ammonia reacts with oxygen it forms nitrogen dioxide and water.
- Now, for balancing this reaction first we need to balance the elements in the reactions other than hydrogen and oxygen atoms.
- In this reaction, only nitrogen is there other than hydrogen and oxygen. First, we will balance the number of nitrogen atoms. The reaction we have:
$NH_{3}+O_{2}\rightarrow NO_{2}+H_{2}O$
- Here, the number of nitrogen atoms is the same on both sides, i.e., one nitrogen atom is present on the reactant and product side. Thus, there is no need to change the reaction.
- Now, balancing the hydrogen atoms in the reaction, the hydrogen atoms in the reactant side is 3 and on the product side is 2. So, multiplying the ammonia with 2 and water with 3, we get:
$2NH_{3}+O_{2}\rightarrow NO_{2}+3H_{2}O$
Now, on the reactants side, the nitrogen is 2 and on the product side, the nitrogen atom is 1. So, multiplying the nitrogen dioxide with 2, we get
$2NH_{3}+O_{2}\rightarrow 2NO_{2}+3H_{2}O$
Now, balancing the oxygen atoms in the reaction, the oxygen atoms on the reactant side is 2 and the number of oxygen atoms on the product side is 7. So, multiplying the oxygen in the reactant with 7 and multiplying both the products with 2, we get
$2NH_{3}+7O_{2}\rightarrow 4NO_{2}+6H_{2}O$
To finally balance this equation, we will multiply ammonia with 2,
$4NH_{3}+7O_{2}\rightarrow 4NO_{2}+6H_{2}O$
Note:
It must be noted that first balance all the elements other than oxygen and hydrogen, and then balance hydrogen, check all the other atoms are balanced then only proceed for balancing of oxygen atoms.
Complete answer:
- The reaction given in the question is the formation of nitrogen dioxide from ammonia when ammonia reacts with oxygen it forms nitrogen dioxide and water.
- Now, for balancing this reaction first we need to balance the elements in the reactions other than hydrogen and oxygen atoms.
- In this reaction, only nitrogen is there other than hydrogen and oxygen. First, we will balance the number of nitrogen atoms. The reaction we have:
$NH_{3}+O_{2}\rightarrow NO_{2}+H_{2}O$
- Here, the number of nitrogen atoms is the same on both sides, i.e., one nitrogen atom is present on the reactant and product side. Thus, there is no need to change the reaction.
- Now, balancing the hydrogen atoms in the reaction, the hydrogen atoms in the reactant side is 3 and on the product side is 2. So, multiplying the ammonia with 2 and water with 3, we get:
$2NH_{3}+O_{2}\rightarrow NO_{2}+3H_{2}O$
Now, on the reactants side, the nitrogen is 2 and on the product side, the nitrogen atom is 1. So, multiplying the nitrogen dioxide with 2, we get
$2NH_{3}+O_{2}\rightarrow 2NO_{2}+3H_{2}O$
Now, balancing the oxygen atoms in the reaction, the oxygen atoms on the reactant side is 2 and the number of oxygen atoms on the product side is 7. So, multiplying the oxygen in the reactant with 7 and multiplying both the products with 2, we get
$2NH_{3}+7O_{2}\rightarrow 4NO_{2}+6H_{2}O$
To finally balance this equation, we will multiply ammonia with 2,
$4NH_{3}+7O_{2}\rightarrow 4NO_{2}+6H_{2}O$
Note:
It must be noted that first balance all the elements other than oxygen and hydrogen, and then balance hydrogen, check all the other atoms are balanced then only proceed for balancing of oxygen atoms.
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