Element	%	Atomic mass	mole ratio	simple ratio
C	38.71	12	$${{38.71} \over {12}} = 3.22$$	$${{3.22} \over {3.22}} = 1$$
H	9.67	1	$${{9.67} \over 1} = 9.67$$	$${{9.67} \over {3.22}} = 1$$
O	51.62	16	$${{51.62} \over {16}} = 3.22$$	$${{3.22} \over {3.22}} = 1$$

Hence empirical formula of the compound would be CH₃O

PbO	+	2HCl	$$\to$$	PbCl2	+	H2O
n mol		2n mol		n mol
$$6.5 \over 224$$ mol		$$3.2 \over 36.5$$ mol
0.029 mol		0.087 mol

Formation of moles of lead(II) chloride depends upon the no. of moles of PbO which acts as a limiting reagent here. So no of moles of PbCl₂ formed will be equal to the no of PbO (i.e 0.029)

Average isotope mass of X

= $${{200 \times 90 + 199 \times 8 + 202 \times 2} \over {90 + 8 + 2}}$$

= $${{18000 + 1592 + 404} \over {100}}$$

= 199.96 a.m.u $$ \approx $$ 200 a.m.u

At STP, 22.4 L H₂

= 6.023 $$ \times $$ 10²³ molecules

15 L H₂ = $${{6.023 \times {{10}^{23}} \times 15} \over {22.4}}$$ = $$4.033 \times {10^{23}}$$

5 L N₂ = $${{6.023 \times {{10}^{23}} \times 5} \over {22.4}}$$ = 1.344 $$ \times $$ 10²³

2 g of H₂ = 6.023 $$ \times $$ 10²³

0.5 g H₂ = $${{6.023 \times {{10}^{23}} \times 0.5} \over {2}}$$ = 1.505 $$ \times $$ 10²³

32 g O₂ = 6.023 $$ \times $$ 10²³

10 g O₂ = $${{6.023 \times {{10}^{23}} \times 10} \over {32}}$$ = 1.882 $$ \times $$ 10²³