Question

How many atoms are present in the given reaction?
$Pb{{(N{{O}_{3}})}_{2}}+2KI\to Pb{{I}_{2}}+2KN{{O}_{3}}$

Answer 1

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.