Chemistry - Stoichiometry Question with Solution | TestHub

(P) ${\mathrm{SO}}_3\Rightarrow$ V.D. $=\frac{\mathrm{MW}}{2}=\frac{80}{2}=40$

No. of electrons in one mole $=40{\mathrm{N}}_{\mathrm{A}}$

(Q) $\mathrm{CO}\Rightarrow \%$ by mass of $\mathrm{C}=\frac{12}{28}\times 100=\frac{300}{7}\%$

$\%$ carbon by moles $=\frac{1}{2}\times 100=50\%$

(R) ${\mathrm{C}}_6{\mathrm{H}}_6\Rightarrow$ Hydrocarbon

$\%$ by mole of $\mathrm{C}=\frac{6}{12}\times 100=50\%$

(S) ${\mathrm{C}}_2{\mathrm{H}}_2\Rightarrow$ Hydrocarbon

$\%$ by mole of $\mathrm{C}=\frac{2}{4}\times 100=50\%$