What volume of air containing 21% of oxygen is required to completely burn 50g of sulfur containing 4% incombustible matter
Mass of sulfur =10050×4=2
Mass of combustible matter=50−2=48= 50-2 = 48=50−2=48
S+O2−−−>SO2S+O_2--->SO_2S+O2−−−>SO2
O2=22.4×4832=33.6O_2= \frac{22.4×48}{32}=33.6O2=3222.4×48=33.6
Volume required=100×33.621=160L=\frac {100×33.6}{21}=160L=21100×33.6=160L
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