Question

	List - I		List - II (At STP)
(A)	$10gCaC{O}_3$ $\underset{decomposition}{\overset{\Delta }{\to }}$	(i)	$0.224LC{O}_2$
(B)	$1.06gN{a}_{2}C{O}_3$ $\stackrel{ExcessHCL}{\to }$	(ii)	$4.48LC{O}_2$
(C)	$2.4gC$ $\underset{combustion}{\overset{Excess{O}_2}{\to }}$	(iii)	$0.448LC{O}_2$
(D)	$0.56gCO$ $\underset{combustion}{\overset{Excess{O}_2}{\to }}$	(iv)	$2.24LC{O}_2$
		(v)	$22.4LC{O}_2$

Match the List I with the List II.

• A. A - iv, B - i, C - ii, D - iii
• B. A - v, B - i, C - ii, D - iii
• C. A - iv, B - i, C - iii, D - ii
• D. A - i, B - iv, C - ii, D - iii

Answer 1

The correct option is A A - iv, B - i, C - ii, D - iii

$\ast$ ${CaCO}_3\to CaO+{CO}_2$

$100g\to 22.4$ litres

$10g\to x$ $x=\frac{22.4\times 10}{100}=2.24$ litres

$\ast$ ${Na}_2{CO}_3+2HCl+H_{2}O+2NaCl+{CO}_2$

$106g⟶22.4$ litres $x=\frac{1.06\times 22.4}{106}=0.224$ lit.

$1.06g⟶x$

$\ast$ $C+O_2\to {CO}_2$

$12\to 22.4$ lit

$2.4\to x$ $x=\frac{2.4\times 22.4}{12}=4.48$ L

$\ast$ $CO+1/2O_2\to {CO}_2$

$28g\to 22.4$ litres

$0.5g\to ?\left(x\right)$ $x=\frac{0.5\times 22.4}{28}=0.448$ lit (approx)