From 18.8 grams of phenol, 52 grams of tribromophenol was obtained. What is the practical yield of this reaction?

Let’s implement the solution:
1. According to the condition of the problem, we compose the equation:
C6H5OH + 3Br2 = C6H2Br3 (OH) + 3HBr – substitutions, tribromophenol was obtained;
2. Calculations:
M (C6H5OH) = 94 g / mol;
M C6H2Br3 (OH) 3 = 330.7 g / mol.
3. Determine the amount of the original substance:
Y (C6H5OH) = m / M = 18.8 / 94 = 0.2 mol;
Y C6H2Br3 (OH) = 0.2 mol since the amount of substances according to the equation is 1 mol.
4. Find the mass and yield of the product:
m (theoretical) = Y * M = 0.2 * 330.7 = 66.14 g;
W = m (practical) / m (theoretical) * 100;
W = 52 / 66.14 * 100 = 78.62%
Answer: the yield of tribromophenol was 78.62%.