To determine the amount of bromine needed to convert 2 g of phenol into 2,4,6-tribromophenol, we follow this procedure:
Step 1: Calculate the molar mass of phenol (C6H5OH)
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Molar mass of phenol = 6(12) + 6(1) + 16 = 94 g/mol
Step 2: Calculate the number of moles of phenol
Number of moles = mass / molar mass = 2 g / 94 g/mol ≈ 0.0213 mol
Step 3: Determine the reaction and molar mass of tribromophenol
Reaction: C6H5OH + 3 Br2 → C6H2Br3OH + 3 HBr
Molar mass of Br2 = 2(80) = 160 g/mol
Step 4: Calculate bromine required for the reaction
Bromine required per mol of phenol = 3 mols of Br2
Number of moles of Br2 = 3 × 0.0213 mol = 0.0639 mol
Mass of Br2 = moles × molar mass = 0.0639 mol × 160 g/mol = 10.224 g
Thus, the amount of bromine required is 10.22 g.
If 0.01 mol of $\mathrm{P_4O_{10}}$ is removed from 0.1 mol, then the remaining molecules of $\mathrm{P_4O_{10}}$ will be: