In an unknown compounds containing $$N$$ and $$H$$
$$\eqalign{ & {\text{given}}\,\,\% \,\,{\text{of}}\,H = 12.5\% \cr & \therefore \,\,\,\% \,\,{\text{of}}\,N = 100 - 12.5 = 87.5\% \cr} $$

Element	Percentage	Atomic ratio	Simple ratio
$$H$$	12.5%	$$\frac{{12.5}}{1} = 12.5$$	$$\frac{{12.5}}{{6.25}} = 2$$
$$N$$	87.5	$$\frac{{87.5}}{{14}} = 6.25$$	$$\frac{{6.25}}{{6.25}} = 1$$

$$\eqalign{ & 2 \times {\text{vapour density}} = Mol.\,wt = mol\,wt. \cr & = 16 \times 2 = 32. \cr & {\text{Molecular formula}} = n \times {\text{empirical formula}} \cr & {\text{mass}}\,\,n = \frac{{32}}{{16}} = 2 \cr & \therefore \,\,{\text{Molecular formula of the compound will be}} \cr & = {\left( {N{H_2}} \right)_2} = {N_2}{H_4} \cr} $$

According to the law of multiple proportions, "if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers."
Fixing the mass of dinitrogen as $$28\,g,$$  masses of dioxygen combined will be 32, 64, 32 and 96 in the given figures. They are in the ratio of 1 : 2 : 1 : 3.

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\,\,\,\,\,Ca{{\left( {OH} \right)}_2}}\limits_{1\,\,mol} \,\,\,\,\,\, + \mathop {\,\,\,N{a_2}S{O_4}}\limits_{1\,\,mol} \,\,\,\,\, \to \mathop {\,\,CaS{O_4}}\limits_{1\,\,mol} + \mathop {2NaOH}\limits_{2\,\,mol} \cr & {\text{Given,}}\,\,\,\,\,\,\,\,200\,\,mmol\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{4}{{142}}mol \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \approx 28\,\,mmol \cr} $$
$${\text{Thus,}}\,\,N{a_2}S{O_4}\,\,{\text{is limiting reagent}}{\text{.}}$$       $${\text{So,}}\,28\,\,mmol\,\,{\text{of}}\,\,CaS{O_4}\,\,{\text{will be produced}}{\text{.}}$$
$$28\,\,mmol\,\,{\text{of}}$$     $$CaS{O_4} \equiv 28 \times {10^{ - 3}} \times 136 = 3.81\,g$$
$$28\,\,mmol\,\,Ca{\left( {OH} \right)_2}\,\,{\text{will produce}}$$       $$ = 56\,\,mmol\,\,{\text{of}}\,\,NaOH.$$
$$\eqalign{ & \left[ {O{H^ - }} \right] = \frac{{56}}{{100}} \times 1000\frac{{mmol}}{L} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 560\,\,mmol/L = 0.56\,\,mol/L \cr} $$

\[\underset{\begin{smallmatrix} 12\times 2+4\times 1 \\ =28\,g \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{4}}}}\,+\underset{\begin{smallmatrix} 16\times 6 \\ =96\,g \end{smallmatrix}}{\mathop{3{{O}_{2}}}}\,\to 2C{{O}_{2}}+2{{H}_{2}}O\]
\[\because \]  For the combustion of $$28 \times {10^{ - 3}}kg$$   of ethylene
oxygen required $$ = 96 \times {10^{ - 3}}kg$$
$$\therefore $$  For the combustion of $$2.8\,kg$$  of ethylene oxygen required
$$ = \frac{{96 \times {{10}^{ - 3}} \times 2.8}}{{28 \times {{10}^{ - 3}}}} = 9.6\,kg$$

$$\eqalign{ & {\text{Number of atoms }} \cr & = {\text{ number of moles}} \times {{\text{N}}_A} \times {\text{atomicity}} \cr & = 0.1 \times 6.023 \times {10^{23}} \times 3 \cr & = 1.806 \times {10^{23}}{\text{atoms}} \cr} $$

$$\eqalign{ & {\text{In}}\,15\,L\,{\text{of}}\,{H_2}\,gas\,at\,STP, \cr & {\text{the number of molecules}} \cr & = \frac{{6.023 \times {{10}^{23}}}}{{22.4}} \times 15 \cr & = 4.033 \times {10^{23}} \cr & {\text{In }}5{\text{ }}L\,{\text{of }}{N_2}{\text{ gas at }}STP, \cr & {\text{the number of molecules}} \cr & = \frac{{6.023 \times {{10}^{23}} \times 5}}{{22.4}} \cr & = 1.344 \times {10^{23}} \cr & {\text{In }}0.5{\text{ }}g{\text{ of }}{H_2}{\text{ gas,}} \cr & {\text{the number of molecules}} \cr & = \frac{{6.023 \times {{10}^{23}} \times 0.5}}{2} \cr & = 1.505 \times {10^{23}} \cr & {\text{In }}10{\text{ }}g{\text{ of }}{O_2}{\text{ gas,}} \cr & {\text{the number of molecules}} \cr & = \frac{{6.023 \times {{10}^{23}} \times 10}}{{32}} \cr & = 1.882 \times {10^{23}} \cr} $$
Hence, maximum number of molecules are present in $$15{\text{ }}L$$  of $${H_2}$$  at $$STP.$$

TIPS/Formulae:
The highest  $$O.S.$$  of an element is equal to the number of its valence electrons
$$\left( {\text{i}} \right){\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }},O.N{\text{.}}\,{\text{of}}\,Fe = + 3,$$       $${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }},O.N\,{\text{of}}\,Co = + 3$$
$$({\text{ii}})\,Cr{O_2}C{l_2},O.N.\,{\text{of}}\,Cr = + 6,$$       $$\left( {{\text{Highest}}\,O.S.\,{\text{of}}\,Cr} \right)$$
$${\left[ {Mn{O_4}} \right]^ - }O.N\,{\text{of}}\,Mn = + 7$$      $$\left( {{\text{Highest}}\;O.S.\,{\text{of}}\,Mn} \right)$$
$$({\text{iii}})\,Ti{O_3},O.N.\,{\text{of}}\,Ti = + 6,$$      $$Mn{O_2}O.N.\,\,{\text{of}}\,\,Mn = + 4$$
$$({\text{iv}})\,{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }},\,O.N.\,\,{\text{of}}\,\,Co$$       $$ = + 3,Mn{O_3},\,O.N.\,\,{\text{of }}\,Mn = + 6$$

$$\eqalign{ & CaC{O_3} + {H_2}S{O_4} \to CaS{O_4} + {H_2}O + C{O_2} \cr & 1\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,100\,g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,22.4\,{\text{litre}} \cr & 1\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10\,g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2.24\,{\text{litre}} \cr} $$

$$\eqalign{ & {\text{No}}{\text{. of moles of}}\,\,NaOH = \frac{{4.28}}{{40}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.107 \cr & {\text{Volume of solution}} = 250\,c{m^3} \cr & M = \frac{n}{{V\left( {{\text{in}}\,\,L} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{0.107}}{{250}} \times 1000 \cr & \,\,\,\,\,\,\, = 0.428\,mol\,{L^{ - 1}} \cr} $$

The following reaction occurs:
$$6F{e^{2 + }} + C{r_2}O_7^{2 - } + 14{H^ + } \to 6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}O$$
From the above equation, we find that Mohr's salt  $$\left( {FeS{O_4}.{{\left( {N{H_4}} \right)}_2}S{O_4}.6{H_2}O} \right)$$      and dichromate reacts in 6 : 1 molar ratio.