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?

Answer 1

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 .