$$\mathop {2KCl{O_3}}\limits_{4.88\,g} \to \mathop {2KCl}\limits_{2.96\,g} + \mathop {3{O_2}}\limits_{1.92\,g} $$
Since, mass of the products is equal to the mass of the reactant, this illustrates the law of conservation of mass.

$${\text{Since}}\,\,1\,\,L = 1000\,\,c{m^3}\,\,$$    $${\text{and}}\,\,1\,\,m = 100\,\,cm$$
$${\text{Hence}},5\,\,L = 5 \times 1000\,c{m^3}$$     $$ = 5 \times {10^{ - 3}}{m^3}$$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C\,\,\,\,\,\,\,\,\,H\,\,\,\,\,\,\,\,\,\,N \cr & {\text{Moles}}\,\,\,\,\,\frac{9}{{12}}\,\,\,\,\,\,\,\,\,\frac{1}{1}\,\,\,\,\,\,\,\,\,\frac{{3.5}}{{14}}\,\,\,\,\,\,\,\,n = \frac{{108}}{{54}} = 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{4}\,\,\,\,:\,\,\,1\,\,\,\,:\,\,\,\,\frac{1}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,:\,\,\,\,\,4\,\,\,\,:\,\,\,\,\,1 \cr & \Rightarrow \,{C_3}{H_4}N \Rightarrow \,M.F. = {\left( {{C_3}{H_4}N} \right)_n} \cr & MF = {\left( {{C_3}{H_4}N} \right)_2} = {C_6}{H_8}{N_2} \cr} $$

$$\eqalign{ & {N_1}{V_1} = {N_2}{V_2} \cr & \left( {{\text{Note}}:{H_3}P{O_3}\,{\text{is}}\,\,{\text{dibasic}}\,\therefore \,M = 2N} \right) \cr & 20 \times 0.2 = 0.1 \times V\,\left( {{\text{Thus,}}\,0.1M = 0.2N} \right) \cr & \therefore V = 40\,ml \cr} $$

TIPS/Formulae :
Sum of oxidation state of all atoms in neutral compound is zero. Let the oxidation state of iron in the complex ion
$$\eqalign{ & {\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {NO} \right)} \right]^{2 + }}:SO_4^{2 - }\,{\text{be}}\,x{\text{;}}\,{\text{then}} \cr & {\text{x + 5}} \times {\text{0 + 0 = + 2}}{\text{.}}\,\,\,\,\therefore x = + 2 \cr} $$

The conditions given are standard conditions
$$224\,mL$$  has mass $$ = 1g;$$
$$22400\,mL$$   will have mass $$ = 100g.$$  This is $$mol.$$  $$wt$$  of gas
$$6.023 \times {10^{23}}$$   molecules have $$3 \times 6.023 \times {10^{23}}$$    atoms since gas is triatomic
∴ weight of one atom
$$\eqalign{ & = \frac{{100}}{{3 \times 6.023 \times {{10}^{23}}}} \cr & = 5.5 \times {10^{ - 23}}g \cr} $$

$$\eqalign{ & {C_2}{H_6} + 3.5{O_2} \to 2C{O_2} + 3{H_2}O\,; \cr & {C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O \cr & {\text{Let volume of ethane is }}x{\text{ litre}} \cr & 22.4 \times 4 = 3.5x + 3\left( {28 - x} \right) \cr & \Rightarrow x = 11.2\,{\text{litre}} \cr & {\text{at constant }}T{\text{ and }}P{\text{,}}\,V \propto n\,; \cr & \Rightarrow {\text{Mole fraction of}}\,{C_2}{H_6}\,{\text{in mixture}} \cr & = \frac{{11.2}}{{28}} \cr & = 0.4 \cr} $$

NOTE :
The sum of oxidation states of all atoms in compound is zero.
Calculation of $$O.S.$$  of $$C$$  in $$C{H_2}O.$$

Writing the equation of combustion of propane $$\left( {{C_3}{H_8}} \right),$$   we get
$$\eqalign{ & {C_3}{H_8} + 5{O_2} \to 3C{O_2} + 4{H_2}O \cr & 1vol\,\,\,\,\,\,\,\,\,\,\,\,\,5vol \cr & 1L\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5L \cr} $$
From the above equation we find that we need 5 $$L$$  of oxygen at $$NTP$$  to completely burn 1 $$L$$  of propane at $$N.T.P.$$
If we change the conditions for both the gases from $$N.T.P.$$  to same conditions of temperature and pressure. The same results are obtained. i.e. 5 $$L$$  is the correct answer.

$$\mathop {2CO}\limits_{2\,vol} + \mathop {{O_2}}\limits_{1\,vol} \to \mathop {2C{O_2}}\limits_{2\,vol} $$
$$1\,vol$$  of $${O_2}$$  reacts with $$2\,vol$$  of $$CO$$
$$1\,L$$  of $${O_2}$$  reacts with $$2\,L$$  of $$CO$$
$$CO$$  left after reaction $$= 3 - 2 = 1 L$$
$$1\,L$$  of $${O_2}$$  produces $$2\,L$$  of $$C{O_2}.$$
Hence, after the reaction, $$CO = 1\,L,C{O_2} = 2\,L$$