How much analyte was in the reaction mixture if 8.25 g of tribromaniline was produced when

How much analyte was in the reaction mixture if 8.25 g of tribromaniline was produced when exposed to bromine water?

Let’s write down the solution:
1. Let’s compose the reaction equation:
C6H5NH2 + 3Br2 = C6H2 (Br) 3NH2 + 3HBr – aniline bromination reaction, tribromaniline precipitate was obtained;
2. Let’s make calculations using the formulas:
M (C6H5NH2) = 12 * 6 + 7 +14 = 93 g / mol;
M C6H2 (Br) 3NH2 = 12 * 6 + 3 * 79.9 + 4 + 14 = 329.7 g / mol;
3. Determine the number of moles of tribromaniline, if the mass is known:
Y C6H5 (Br) 3NH2 = m / M = 8.25 / 329.7 = 0.025 mol;
4. Let’s make the proportion:
X mol (C6H5NH2) – 0.025 mol C6H5 (Br) 3NH2;
-1 mol -1 mol hence, X mol (C6H5NH2) = 0.025 * 1/1 = 0.025 mol;
5. Let’s calculate the mass of aniline by the formula:
M (C6H5NH2) = Y * M = 0.025 * 93 = 2.3 g.
Answer: aniline weighing 2.3 g is required to carry out the reaction.