Question

A sample of ${\mathrm{CaCO}}_3$ is $50\%$ pure. On heating this $1.12\mathrm{L}$ of ${\mathrm{CO}}_2$ (at STP) is obtained. What is the total amount of residue left (assuming non-volatile impurity)?

• A. $2.8\mathrm{g}$
• B. $3.8\mathrm{g}$
• C. $7.8\mathrm{g}$
• D. $8.9\mathrm{g}$

Answer 1

The correct option is C

$7.8\mathrm{g}$

The correct option is (C):

Explanation for the correct answer:

Step 1: Finding the number of moles of ${\mathrm{CaCO}}_3$ reacted:

As, the number of moles of ${\mathrm{CO}}_2$ evolved is: $\frac{1.12}{22.4}=0.05\mathrm{moles}$

Reaction : $\underset{\mathrm{Calcium}\mathrm{carbonate}}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)}\stackrel{∆}{\to }\underset{\mathrm{Calcium}\mathrm{oxide}}{\mathrm{CaO}\left(\mathrm{s}\right)\downarrow }\underset{\mathrm{Carbon}\mathrm{dioxide}}{+{\mathrm{CO}}_2\left(\mathrm{g}\right)\uparrow }$

So, the moles of ${\mathrm{CaCO}}_3$ reacted is : $0.05\mathrm{moles}$

Step 2: Finding the total amount of the residue left:

Mass = $0.05\times 100=5\mathrm{gm}$ = $50\%\mathrm{of}{\mathrm{CaCO}}_3\mathrm{sample}$

Total weight= $2\times 5=10\mathrm{gm}$

Therefore, the residue left by ${\mathrm{CaCO}}_3$ : $5\mathrm{gm}$

And residue left by $\mathrm{CaO}$: $56\times 0.05=2.8\mathrm{gm}$

So, the total residue left would be: $5+2.8=7.8\mathrm{gm}$

Hence, the correct option is (C): $7.8\mathrm{g}$