X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X = g. (nearest integer) [Given: molar mass (in g mol^-

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]

Accepted Answer

Given: Mass of m-dinitrobenzene obtained = $4.2 \, \text{g}$ .

Find: Mass $X$ of nitrobenzene required.

The reaction 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}

So, nitrobenzene and m-dinitrobenzene are in a $1:1$ mole ratio.

Molar mass of nitrobenzene =

(6 \times 12) + (5 \times 1) + (1 \times 14) + (2 \times 16) = 123 \, \text{g/mol}

Molar mass of m-dinitrobenzene =

(6 \times 12) + (4 \times 1) + (2 \times 14) + (4 \times 16) = 168 \, \text{g/mol}

Moles of m-dinitrobenzene produced =

\frac{4.2}{168} = 0.025 \, \text{mol}

Using the $1:1$ stoichiometric ratio, moles of nitrobenzene reacted = $0.025 \, \text{mol}$ .

Therefore, mass of nitrobenzene =

0.025 \times 123 = 3.075 \, \text{g}

The nearest integer is $3$ . Therefore, $X = 3 \, \text{g}$ .

Given: Product mass = 4.2 g of m-dinitrobenzene. Find: Initial mass X of nitrobenzene. First calculate the molar masses from the given atomic masses: - Nitrobenzene, C6 H5 NO2 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123 g/mol - m-dinitrobenzene, C6 H4 N2 O4 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168 g/mol From the balanced equation, C6 H5 NO2 + HNO3 C6 H4 N2 O4 + H2 O 1 mole of nitrobenzene gives 1 mole of m-dinitrobenzene. Now convert the given product mass into moles: moles of m-dinitrobenzene = 4.2168 = 0.025 Hence, moles of nitrobenzene consumed = 0.025. Now convert moles of nitrobenzene into mass: X = 0.025 123 = 3.075 g Rounding to the nearest integer gives 3. So the numerical answer is 3.

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