A garderner fugimates his greenhouse (measuring 2m x 3m x 4m) by burning a sulphur "candle". A gaseous concentration of sulphur dioxide of 50ppm (parts per million) by volume is found to be effective in getting rid of pests and moulds.
How many grams of sulphur must the garderner burn in order to produce a concentration of 50ppm of SO2 in his greenhouse?
[Vol of 1 mol of gas = 24 dm³ (o.o24dm³) at r.t.p]
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Answer:
vol of greenhouse = 2 x 3 x 4
= 24 m³
vol of SO2 = (50/106) x 24 = 1.2 x 10-3 m³
mol of SO2 needed = (1.2 x 10-3) / 0.024 = 0.050
Since S + O2 -> SO2 ,
mol of S = mol of SO2 = 0.050
therefore,
mass of S = 0.050 x 32.1
= 1.61g (3s.f.)
