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.

Answer 2

To solve the problem, we need to determine the mass (X g) of nitrobenzene that produces 4.2 g of m-dinitrobenzene through nitration.

First, calculate the molar masses: 
C₆H₅NO₂ (nitrobenzene):
Molar mass = 12×6 (C) + 1×5 (H) + 14×1 (N) + 16×2 (O) = 123 g mol^-1
C₆H₄N₂O₄ (m-dinitrobenzene):
Molar mass = 12×6 + 1×4 + 14×2 + 16×4 = 168 g mol^-1

From the balanced chemical equation:
C₆H₅NO₂ + HNO₃ → C₆H₄N₂O₄ + H₂O
Moles of nitrobenzene = Moles of m-dinitrobenzene

Calculate moles from the given mass of m-dinitrobenzene:
moles = 4.2 g / 168 g mol^-1 = 0.025 moles

Since the reaction is 1:1, moles of nitrobenzene = 0.025 moles

Calculate mass of nitrobenzene (X g):
X = moles × molar mass = 0.025 × 123 = 3.075 g

Nearest integer: X = 3 g

Confirming the range: X = 3 falls within the range (3, 3).