What volume of air at 1 atm and 273 K containing 2118S8s-O2g- SO2g

Weight of S_8 in sample =160 g Moles of S_8=16032× 8=0.625 Number of moles of O_2 required =0.625× 8 Volume of O_2 required at STP=22.4× 5 ∴ nbsp; Volume of ...

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Standard XIII

Chemistry

Mole Concept

What volume o...

Question

What volume of air at $1 a t m$ and $273 K$ containing 21% of oxygen by volume is required to completely burn sulphur $(S_{8})$ present in $200 g$ of sample, which contains $20 %$ inert material which does not burn. Sulpur burns according to the reaction
$\frac{1}{8} S_{8} (s) + O_{2} (g) \to S O_{2} (g)$

23.52 L

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320 L

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112 L

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533.33 L

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Solution

The correct option is D $533.33 L$
$Weight of S_{8} in sample = 160 g$
$Moles of S_{8} = \frac{160}{32 \times 8} = 0.625$
$Number of moles of O_{2} required = 0.625 \times 8$
$Volume of O_{2} required at S T P = 22.4 \times 5$
$∴ Volume of air required at S T P = 22.4 \times 5 \times \frac{100}{21} = 533.33 L$

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of a sample, which contains 20% inert material that does not burn? Sulphur burns according to the reaction:

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present in

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What volume of air at 1 atm and 273 K containing 21% of oxygen by volume is required to completely burn sulphur (S8) present in 200 g of sample, which contains 20% inert material which does not burn. Sulpur burns according to the reaction 18S8(s)+O2(g)→SO2(g)

The correct option is D 533.33 LWeight of S8 in sample=160 g Moles of S8=16032×8=0.625 Number of moles of O2 required=0.625×8 Volume of O2 required at STP=22.4×5 ∴Volume of air required at STP=22.4×5×10021=533.33 L