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]
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]
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Correct Answer: 3
Approach Solution - 1
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.
Approach Solution -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).
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