Question

9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/ conversion) is ________g. (nearest integer)

Answer 1

The reaction is as follows:
$\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \xrightarrow{T<5^\circ \text{C}} \text{C}_6\text{H}_5\text{-N}_2^+\text{Cl}^-\xrightarrow{\text{Phenol}} \text{Orange Dye}$
1 mole of aniline (C$_6$H$_5$NH$_2$) produces 1 mole of orange dye.
Molar mass of aniline = 93 g mol$^{-1}$.
Molar mass of orange dye = 199 g mol$^{-1}$.
Moles of aniline used:
$\text{moles of aniline} = \frac{\text{mass of aniline}}{\text{molar mass of aniline}} = \frac{9.3}{93} = 0.1 \, \text{mol}.$

Mass of orange dye produced:
$\text{mass of orange dye} = \text{moles of aniline} \times \text{molar mass of orange dye} = 0.1 \times 199 = 19.9 \, \text{g} \approx 20 \,\text{g}.$

Answer 2

The reaction is as follows:
$\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \xrightarrow{T<5^\circ \text{C}} \text{C}_6\text{H}_5\text{-N}_2^+\text{Cl}^-\xrightarrow{\text{Phenol}} \text{Orange Dye}$
1 mole of aniline (C$_6$H$_5$NH$_2$) produces 1 mole of orange dye.
Molar mass of aniline = 93 g mol$^{-1}$.
Molar mass of orange dye = 199 g mol$^{-1}$.
Moles of aniline used:
$\text{moles of aniline} = \frac{\text{mass of aniline}}{\text{molar mass of aniline}} = \frac{9.3}{93} = 0.1 \, \text{mol}.$

Mass of orange dye produced:
$\text{mass of orange dye} = \text{moles of aniline} \times \text{molar mass of orange dye} = 0.1 \times 199 = 19.9 \, \text{g} \approx 20 \,\text{g}.$