Question

X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X =_____ g. (nearest integer) [Given : molar mass (in g mol\(^{-1}\)) C : 12, H : 1, O : 16, N : 14]

Hint:

In stoichiometric calculations, always start by calculating the number of moles of the given substance using its mass and molar mass. Then, use the mole ratio from the balanced chemical equation to find the moles of the required substance. Finally, convert the moles of the required substance back to mass using its molar mass.

Answer 1

Correct Answer: 3

The nitration of nitrobenzene to m-dinitrobenzene proceeds via the following reaction: \[ \text{C}_6\text{H}_5\text{NO}_2 + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4, \Delta} \text{C}_6\text{H}_4(\text{NO}_2)_2 + \text{H}_2\text{O} \] The molar mass (MW) of nitrobenzene (C\( _6 \)H\( _5 \)NO\( _2 \)) is calculated as: \[ (6 \times 12) + (5 \times 1) + (1 \times 14) + (2 \times 16) = 72 + 5 + 14 + 32 = 123 \, \text{g/mol} \] The molar mass (MW) of m-dinitrobenzene (C\( _6 \)H\( _4 \)N\( _2 \)O\( _4 \)) is calculated as: \[ (6 \times 12) + (4 \times 1) + (2 \times 14) + (4 \times 16) = 72 + 4 + 28 + 64 = 168 \, \text{g/mol} \] According to the reaction stoichiometry, one mole of nitrobenzene yields one mole of m-dinitrobenzene.
The moles of m-dinitrobenzene produced are determined by: \( \frac{\text{mass of m-dinitrobenzene}}{\text{molar mass of m-dinitrobenzene}} \) \[ \text{Moles of m-dinitrobenzene} = \frac{4.2 \, \text{g}}{168 \, \text{g/mol}} = 0.025 \, \text{mol} \] Given the 1:1 mole ratio between nitrobenzene and m-dinitrobenzene, 0.025 moles of nitrobenzene were reacted. 
The mass of nitrobenzene reacted (X) is calculated as: moles of nitrobenzene × molar mass of nitrobenzene \[ X = 0.025 \, \text{mol} \times 123 \, \text{g/mol} = 3.075 \, \text{g} \] The nearest integer to 3.075 is 3. Thus, X = 3 g.