$$\eqalign{ & {M_1}{V_1} + {M_2}{V_2} = MV \cr & 0.5 \times 50 + 0.25 \times 150 = M \times 250 \cr & M = \frac{{25 + 37.5}}{{250}} \cr & \,\,\,\,\,\,\,\, = 0.25\,M \cr} $$

$$\eqalign{ & {\text{Atomic mass of an element}} \cr & = {\text{Mass of one atom}} \times {N_A} \cr & = 3.32 \times {10^{ - 23}} \times 6.023 \times {10^{23}} \cr & = 19.99 \approx 20\,g \cr & {\text{No}}{\text{. of gram atoms}} \cr & = \frac{{{\text{Mass of element in gram}}}}{{{\text{Atomic mass in gram}}}} \cr & = \frac{{20 \times 1000}}{{20}} \cr & = 1000 \cr} $$

$$\eqalign{ & {\text{Mass of oxide = Mass of metal + Mass of oxygen}} \cr & 2.74 = 1.53 + {W_{{\text{Oxygen}}}} \Rightarrow {W_{{\text{Oxygen}}}} = 1.21\,g \cr & {\text{Moles of}}\,\,V = \frac{{1.53}}{{51}} = 0.03 \cr & {\text{Moles of}}\,\,{\text{O = }}\frac{{1.21}}{{16}} = 0.075 \cr & {V_{0.03}}\,{O_{0.075}} \cr & V\,{O_{2.5}} \Rightarrow {V_2}{O_5} \cr} $$

Let $$wt.$$  of $$N{H_4}N{O_3}$$   and $${\left( {N{H_4}} \right)_2}HP{O_4}$$    are $$x$$ and $$y$$ gram respectively
$$\eqalign{ & \frac{{\frac{x}{{80}} \times 2 \times 14 + \frac{y}{{132}} \times 2 \times 14}}{{x + y}} \times 100 = 30.4 \cr & \Rightarrow x:y = 2:1 \cr} $$

$$\mathop {4N{H_3}\left( g \right)}\limits_{4\,mol} + \mathop {5{O_2}\left( g \right)}\limits_{5\,mol} \to \mathop {4NO\left( g \right)}\limits_{4\,mol} + \mathop {6{H_2}O\left( l \right)}\limits_{6\,mol} $$
According to equation,
$$5{\text{ }}moles$$   of $${O_2}$$  required $$ = 4{\text{ }}moles$$   of $$N{H_3}$$
$$1{\text{ }}mole$$   of $${O_2}$$  requires $$ = \frac{4}{5} = 0.8\,mole$$    of $$N{H_3}$$
While $$1{\text{ }}mole$$   of $$N{H_3}$$  requires $$ = \frac{5}{4} = 1.25\,moles$$     of $${O_2}$$
As there is $$1{\text{ }}mole$$   of $$N{H_3}$$  and $$1{\text{ }}mole$$   of $${O_2}{\text{,}}$$  so all the oxygen will be consumed.

$$\eqalign{ & \left( {\text{A}} \right):28\,g\,\,{\text{of}}\,\,He = \frac{{28}}{4} = 7\,mol \cr & \left( {\text{B}} \right):46\,g\,\,{\text{of}}\,\,Na = \frac{{46}}{{23}} = 2\,mol \cr & \left( {\text{C}} \right):60\,g\,\,{\text{of}}\,\,Ca = \frac{{60}}{{40}} = 1.5\,mol \cr & \left( {\text{D}} \right):27\,g\,\,{\text{of}}\,\,Al = \frac{{27}}{{27}} = 1\,mol \cr} $$

$$0.5\% $$  by weight means if $$Mol.$$  $$wt.$$  is 100 then mass of $$Se$$  is 0.5. If at least one atom of $$Se$$  is present in the molecule then
$$\eqalign{ & M.\,\,Wt = \frac{{100}}{{0.5}} \times 78.4 \cr & = 1.568 \times {10^4} \cr} $$

\[\underset{\begin{smallmatrix} 1\,V \\ 10\,L \end{smallmatrix}}{\mathop{{{N}_{2}}}}\,+\underset{\begin{smallmatrix} 3\,V \\ 30\,L \end{smallmatrix}}{\mathop{3{{H}_{2}}}}\,\to \underset{\begin{smallmatrix} 2\,V \\ 20\,L \end{smallmatrix}}{\mathop{2N{{H}_{3}}}}\,\]
As only $$50\% $$  of the expected product is formed, hence only $$10 L$$  of \[N{{H}_{3}}\]  is formed.
Thus, for the production of $$10 L$$  of \[N{{H}_{3}}\]  , $$5L$$  of $${N_2}$$  and $$15L$$  of $${H_2}$$  are used and composition of gaseous mixture under the aforesaid condition in the end is
$$\eqalign{ & {H_2} = 30 - 15 = 15\,L \cr & {N_2} = 30 - 5 = 25\,L \cr & N{H_3} = 10L \cr} $$

2.5 × 1.25 = 3.125
Since 2.5 has two significant figures, the result should not have more than two significant figures. Hence, the answer will be 3.1.

$$\eqalign{ & CaC{O_3} \to CaO + C{O_2} \cr & 20\,g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8.8\,g\,\,\,\,\,\,\,\,\,11.2\,g \cr} $$
mass of reactant $$ = $$ mass of products $$ = 20g.$$
Hence the law of conservation of mass is obeyed.