Question

10 g mixture of NaHCO$_3$ and Na$_2$CO$_3$ has 1.68 g NaHCO$_3$. It is heated at 400 K. Weight of the residue will be:
(A)– 9.38 g
(B)– 8.32 g
(C)– 10.0 g
(D)– 1.68 g

Answer 1

Hint: First, we will write the balanced chemical reaction decomposition of sodium bicarbonate. Calculate their molar masses, and weight decomposed; when the reaction occurred. The weight of residue can be calculated.

Complete step by step answer:
- First, let us write the decomposition reaction of sodium bicarbonate.
- By writing the balanced chemical reaction we will see that on heating it will be decomposed into sodium carbonate, and water; and the carbon dioxide gas is also evolved. For this, stoichiometric coefficient should be taken into account.
-Stoichiometric coefficients is nothing but the number of atoms/ions or molecules that participate in the reaction. These are used to make sure whether the equation is balanced or not.
-So, the balanced chemical reaction is
\[\begin{matrix}
   2NaHC{{O}_{3}} \\
   168 \\
\end{matrix}+\text{ Heat}\to \begin{matrix}
   N{{a}_{2}}C{{O}_{3}} \\
   106 \\
\end{matrix}+\begin{matrix}
   {{H}_{2}}O \\
   18 \\
\end{matrix}+\begin{matrix}
   C{{O}_{2}} \\
   44 \\
\end{matrix}\]

-Now, we will calculate the molar mass of $NaHC{{O}_{3}}$ i.e. 23 + 1 + 12 + 3(16) = 84 g/ mol.
-Similarly, the mass of carbon-dioxide is 44 g/mol, and water is 18 g/mol.
-Now, according to the reaction there are 2 moles of sodium bicarbonate getting decomposed, so it can be written as 284 = 168 g of sodium bicarbonate.
-It is decomposed into carbon-dioxide, and water as mentioned, so 44 + 18 = 62 g

The above numerical can be solved using UNITARY METHOD.
Here, 2 moles of sodium bicarbonate gives 1 mole of ${{H}_{2}}O$ and 1 mole of $C{{O}_{2}}$ respectively.
Writing in terms of molar mass, we get
$\because $ 168g of sodium bicarbonate when subjected to heat gives off carbon-dioxide and water i.e. 62g.
$\therefore $ 1g of sodium bicarbonate when subjected to heat gives off carbon-dioxide and water i.e. $\dfrac{62}{168}g$
This implies that
1.68g of sodium bicarbonate is decomposed to $\dfrac{62}{168}\text{x1}\text{.68=0}\text{.62}$g of carbon-dioxide and water
Then, the weight of residue = Total weight of the mixture – Combined weight of $C{{O}_{2}}\text{ and }{{H}_{2}}O$
=10 – 0.62 g = 9.38 g


Hence, the correct option is (A).


Note: Don’t get confused while calculating the weight of residue. We will consider the sodium bicarbonate, as it acts as a reactant getting decomposed. The temperature is given for the knowledge, otherwise it is not required in the calculation.