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 reaction for the nitration of nitrobenzene to m-dinitrobenzene is: \[ \text{C}_6\text{H}_5\text{NO}_2 + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4, \Delta} \text{C}_6\text{H}_4(\text{NO}_2)_2 + \text{H}_2\text{O} \] Nitrobenzene (C\( _6 \)H\( _5 \)NO\( _2 \)) has a molar mass (MW) of: \[ (6 \times 12) + (5 \times 1) + (1 \times 14) + (2 \times 16) = 72 + 5 + 14 + 32 = 123 \, \text{g/mol} \] m-dinitrobenzene (C\( _6 \)H\( _4 \)N\( _2 \)O\( _4 \)) has a molar mass (MW) of: \[ (6 \times 12) + (4 \times 1) + (2 \times 14) + (4 \times 16) = 72 + 4 + 28 + 64 = 168 \, \text{g/mol} \] From the stoichiometry of the reaction, 1 mole of nitrobenzene produces 1 mole of m-dinitrobenzene. 
Moles of m-dinitrobenzene produced = \( \frac{\text{mass of m-dinitrobenzene}}{\text{molar mass of m-dinitrobenzene}} \) \[ \text{Moles of m-dinitrobenzene} = \frac{4.2 \, \text{g}}{168 \, \text{g/mol}} = 0.025 \, \text{mol} \] Since the mole ratio of nitrobenzene to m-dinitrobenzene is 1:1, the moles of nitrobenzene reacted are also 0.025 mol. 
Mass of nitrobenzene reacted (X) = 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. Therefore, X = 3 g.