9.3 g of aniline is converted into 1,3,5-tribromobenzene in three steps: bromination with $Br X_{2}$ water to 2,4,6-tribromoaniline (90 per cent yield), diazotisation (80 per cent yield) and reduction of the diazonium salt with $H X_{3} PO X_{2}$ (75 per cent yield). Atomic masses: H = 1, C = 12, N = 14, Br = 80. Calculate the mass in grams of 1,3,5-tribromobenzene obtained, to one decimal place.

Numerical answer type, tolerance 0.1.

Answer: 17

Moles of aniline (

C X_{6} H X_{5} NH X_{2}

, M = 93) = 9.3/93 = 0.100 mol. Each step is 1:1, so the overall yield is 0.90 x 0.80 x 0.75 = 0.54 and the moles of product = 0.100 x 0.54 = 0.054. The molar mass of

C X_{6} H X_{3} Br X_{3}

= 72 + 3 + 240 = 315 g/mol, so the mass = 0.054 x 315 = 17.01 g, i.e. 17.0 g.

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Answer: 17

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