Calculate volume of O2 at NTP required for the complete burning of 12 gram of Sulphur?

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Solution

S + O2 ---> SO2
According to this reaction 32 g of sulphur reacts with 32 g of oxygen.
So ,
12 g of sulphur reacts with 12 g of oxygen.
Also at STP volume of 32 g oxygen is 22.4 L .
So volume of 12 g of oxygen is. (22.4 ÷ 32)× 12 = 8.4 L

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Calculate volume of O2 at NTP required for the complete burning of 12 gram of Sulphur?

S + O2 ---> SO2 According to this reaction 32 g of sulphur reacts with 32 g of oxygen. So , 12 g of sulphur reacts with 12 g of oxygen. Also at STP volume of 32 g oxygen is 22.4 L . So volume of 12 g of oxygen is. (22.4 ÷ 32)× 12 = 8.4 L

S + O2 ---> SO2
According to this reaction 32 g of sulphur reacts with 32 g of oxygen.
So ,
12 g of sulphur reacts with 12 g of oxygen.
Also at STP volume of 32 g oxygen is 22.4 L .
So volume of 12 g of oxygen is. (22.4 ÷ 32)× 12 = 8.4 L