
How many atoms are present in the given reaction?
$ Pb{{(N{{O}_{3}})}_{2}}+2KI\to Pb{{I}_{2}}+2KN{{O}_{3}} $
Answer
503.1k+ views
Hint :To calculate or to find the number of atoms in a reaction simply multiply all the subscripts in the molecule by the coefficient. This will give you atoms present in each element or molecule, then add all the atoms you calculated for each reactant and product in a chemical equation and you will get the desired answer.
Complete Step By Step Answer:
In any chemical equation there are two things happening at a time, one reactant reacting and second products forming due to reaction. We first count atoms present in the reactant side of the given equation.
$ Pb{{(N{{O}_{3}})}_{2}} $ here we have one atom of lead, two atoms of nitrogen and six atoms of oxygen. Total number of atoms here is $ 9 $ .
$ KI $ here we have one atom of potassium and one atom of iodine, thereby giving a total of $ 2 $ atoms. But here we have to multiply this number by two since this is the coefficient (refer the hint) therefore net atoms are $ 4 $ .
Now let’s count the total number of atoms present in the product side of a given chemical equation.
$ Pb{{I}_{2}} $ here we have one atom of lead and two atoms of iodine, giving us a total of $ 3 $ atoms.
$ KN{{O}_{3}} $ Here we have one atom of potassium, one atom of nitrogen and three atoms of oxygen, giving us a total of $ 5 $ atoms. Now we have to multiply it by two in order to satisfy the equation hence total atoms now become $ 10 $ .
Total number of atoms present in the given equation is
$ 9+4+3+10=26 $
$ 26 $ atoms are present in the given reaction.
Note :
Whenever you are asked to find the number of atoms present in a chemical reaction first check whether the reaction is balanced or not, if not then first balance the reaction and then follow the procedure as described above.
Complete Step By Step Answer:
In any chemical equation there are two things happening at a time, one reactant reacting and second products forming due to reaction. We first count atoms present in the reactant side of the given equation.
$ Pb{{(N{{O}_{3}})}_{2}} $ here we have one atom of lead, two atoms of nitrogen and six atoms of oxygen. Total number of atoms here is $ 9 $ .
$ KI $ here we have one atom of potassium and one atom of iodine, thereby giving a total of $ 2 $ atoms. But here we have to multiply this number by two since this is the coefficient (refer the hint) therefore net atoms are $ 4 $ .
Now let’s count the total number of atoms present in the product side of a given chemical equation.
$ Pb{{I}_{2}} $ here we have one atom of lead and two atoms of iodine, giving us a total of $ 3 $ atoms.
$ KN{{O}_{3}} $ Here we have one atom of potassium, one atom of nitrogen and three atoms of oxygen, giving us a total of $ 5 $ atoms. Now we have to multiply it by two in order to satisfy the equation hence total atoms now become $ 10 $ .
Total number of atoms present in the given equation is
$ 9+4+3+10=26 $
$ 26 $ atoms are present in the given reaction.
Note :
Whenever you are asked to find the number of atoms present in a chemical reaction first check whether the reaction is balanced or not, if not then first balance the reaction and then follow the procedure as described above.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

