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# After balancing the reaction find the value of $x$ in $Pb{(N{O_3})_2} + \Delta \to PbO + xN{O_2} + {O_2}$ (A) $4$(B) $3$(C) $5$(D) $2$

Last updated date: 18th Sep 2024
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Hint: In a chemical reaction we do not change the quantity of each element which means that the number of atoms on both sides of the equation should be equal. Thus, if we try to change the scalar quantity for each molecular formula, we might get a balanced equation. So, we will use the try and hit method in this case.

Complete step-by-step solution:To find the value of $x$ we need to first balance the equation and for that we need to check if the number of atoms on both the sides of the reaction is equal or not.

 Number of atoms in reactants Number of atoms in products $Pb = 1 \\ N = 2 \\ O = 6 \\$ $Pb = 1 \\ N = 1 \\ O = 5 \\$

Hence it is clear from the above table that the number of atoms are not equal on both side of the equation.
Now we will try to balance the equation:
$Pb{(N{O_3})_2} + \Delta \to PbO + N{O_2} + {O_2}$

 $Pb{(N{O_3})_2}$ $+$ $\to$ $PbO$ $+$ $N{O_2}$ $+$ ${O_2}$ $Pb = 1 \times 2 = 2 \\ N = 2 \times 2 = 4 \\ O = 6 \times 2 = 12 \\$ $Pb = 1 \times 2 = 2 \\ O = 1 \times 2 = 2 \\$ $N = 1 \times 4 = 4 \\ O = 2 \times 4 = 8 \\$ ${O_2} = 2$

The atoms on both sides are now balanced.

We will hence place $2$ before $Pb{(N{O_3})_2}$ and $PbO$ to balance the $Pb$ and oxygen atoms and in last to balance the nitrogen atoms we will put $4$ before $N{O_2}$ .
Hence the correctly balanced equation will be: -
$2Pb{(N{O_3})_2} + \Delta \to 2PbO + 4N{O_2} + {O_2}$
So, the value of $x$ will be equal to $4$ .

Thus, the correct option is Option A.

Additional information: In this reaction thermal decomposition of lead nitrate is taking place when it is being heated and $N{O_2}$ gas is being evolved which appears as brown fumes. This $N{O_2}$ gas undergoes dimerization when it is cooled to form a stable ${N_2}{O_4}$ molecule.

Note:To balance an equation you must change the coefficients not the subscripts in the formula. You never change numbers that are already a part of the chemical reaction. Always balance polyatomic ions as a whole. Coefficients do not go in the middle of a molecule; they always must go in front of the entire molecule.