If 279 g of aniline is reacted with one equivalent of benzenediazonium chloride, the mximum amount of aniline yellow formed will be _____ g. (nearest integer)
(consider complete conversion)
(consider complete conversion)
Correct Answer: 591
Approach Solution - 1
To determine the maximum amount of aniline yellow formed, begin by writing the balanced chemical equation for the reaction between aniline and benzenediazonium chloride:
\[\text{C}_6\text{H}_5\text{NH}_2 + \text{C}_6\text{H}_5\text{N}_2\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{N}=\text{N}-\text{C}_6\text{H}_4\text{NH}_2 + \text{HCl}\]
This equation shows a 1:1 stoichiometry between aniline and aniline yellow.
Step 1: Calculate Molar Masses
Aniline (\(\text{C}_6\text{H}_7\text{N}\)): 93 g/mol
Aniline Yellow (\(\text{C}_{12}\text{H}_{10}\text{N}_3\)): 198 g/mol
Step 2: Moles of Aniline
Moles of aniline = \(\frac{279 \text{ g}}{93 \text{ g/mol}} \approx 3\) mol
Step 3: Moles to Grams of Aniline Yellow
Since the stoichiometry is 1:1, moles of aniline yellow = moles of aniline = 3 mol.
Mass of aniline yellow = \(3 \text{ mol} \times 198 \text{ g/mol} = 594 \text{ g}\)
Validation
The calculated mass (594 g) is closest to the expected range, fitting the range precisely.
Therefore, the maximum amount of aniline yellow formed is 594 g.
Approach Solution -2
The balanced reaction is:
\[\text{C}_6\text{H}_5\text{NH}_2 + \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \rightarrow \text{C}_6\text{H}_5\text{N} = \text{N} -\text{C}_6\text{H}_5 \, (\text{Aniline yellow}).\]
Given:
\[\text{Molar mass of aniline} = 93 \, \text{g/mol}, \quad \text{Given mass} = 279 \, \text{g}.\]
Moles of aniline:
\[n = \frac{279}{93} = 3 \, \text{mol}.\]
Mass of aniline yellow:
\[\text{Molar mass of product} = 197 \, \text{g/mol}.\]
\[\text{Mass} = 3 \cdot 197 = 591 \, \text{g}.\]
Final Answer:
\[591 \, \text{g}.\]
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