Some Basic Concepts of Chemistry 2 Question 50
49. $4.08 \mathrm{~g}$ of a mixture of $\mathrm{BaO}$ and unknown carbonate $\mathrm{MCO}_{3}$ was heated strongly. The residue weighed $3.64 \mathrm{~g}$. This was dissolved in $100 \mathrm{~mL}$ of $1 \mathrm{~N} \mathrm{HCl}$. The excess acid required $16 \mathrm{~mL}$ of $2.5 \mathrm{~N} \mathrm{NaOH}$ solution for complete neutralisation. Identify the metal $M$.
$(1983,4 \mathrm{M})$
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Solution:
- During heating $M \mathrm{CO}{3}$ is converted into $M \mathrm{O}$ liberating $\mathrm{CO}{2}$ while $\mathrm{BaO}$ is remaining unreacted :
$M \mathrm{CO}{3}(s) \xrightarrow{\text { Heat }} M \mathrm{O}(s)+\mathrm{CO}{2}(g) \uparrow \quad 0.44 \mathrm{~g}=0.01 \mathrm{~mol}$
$$ \frac{\mathrm{BaO}(s)}{4.08 \mathrm{~g}} \quad \frac{\mathrm{BaO}(s)}{3.64 \mathrm{~g}} $$
From the decomposition information, it can be deduced that the original mixture contained 0.01 mole of $M \mathrm{CO}_{3}$ and the solid residue, obtained after heating, contain 0.01 mole ( 10 millimol) of $M \mathrm{O}$.
Also, millimol of $\mathrm{HCl}$ taken initially $=100$
millimol of $\mathrm{NaOH}$ used in back-titration $=16 \times 2.5=40$
$\Rightarrow$ millimol of $\mathrm{HCl}$ reacted with oxide residue $=60$
$\mathrm{HCl}$ reacts with oxides as :
$$ \begin{aligned} & \underset{10 \text { millimol }}{M O}+\underset{20 \text { millimol }}{2 \mathrm{HCl}} \longrightarrow M_{2}+\mathrm{H}{2} \mathrm{O} \ & \mathrm{BaO}+2 \mathrm{HCl} \longrightarrow \mathrm{BaCl}{2}+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$
$60-20=40$ millimol
Therefore, the residue contain 20 millimol of $\mathrm{BaO}$.
Also, $\quad$ molar mass of $\mathrm{BaO}=138+16$
$\Rightarrow \quad$ Mass of $\mathrm{BaO}=\frac{154 \times 20}{1000}=3.08 \mathrm{~g}$
$\Rightarrow$ Mass of $M \mathrm{CO}_{3}=4.08-3.08=1.0 \mathrm{~g}$
$\because \quad 0.01$ mole of $M \mathrm{CO}_{3}$ weight $1.0 \mathrm{~g}$
$\therefore 1$ mole of $M \mathrm{CO}_{3}=100 \mathrm{~g}$
$\Rightarrow 100=($ Atomic weight of metal $)+(12+3 \times 16)$
$\Rightarrow$ Atomic weight of metal $=40$, i.e. $\mathrm{Ca}$ Download Chapter Test
