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Question
What volume of hydrogen sulphide at STP will burn in oxygen to yield 12.8g what volume of oxygen would be required for complete combustion?
Solution
From the equation we know that 2 moles of H2S burn in presence of 3 moles of Oxygen so :
44.8L of H2S requires = 67.2L of oxygen
4.48L of H2S will require = 67.2/44.8 x 4.48 = 6.72L
So, 6.72L of oxygen would be required for complete combustion .
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