what is the volume of oxygen at STP required for complete combustion of 32 grams of ch4
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CH4 + 2O2 = CO2 + 2 H2O
Mass of CH4 = 32 g; Molecular mass of CH4 = 16
Moles of CH4 = 32/16 = 2
1 mole of CH4 need 2 moles of Oxygen ( from equation)
2 moles of CH4 need 4 moles of oxygen.
1 mole of oxygen at stp occupies a volume of 22.4 litres
4 moles of oxygen at stp will have a volume of 4 x 22.4 = 89.6 litres.
Mass of CH4 = 32 g; Molecular mass of CH4 = 16
Moles of CH4 = 32/16 = 2
1 mole of CH4 need 2 moles of Oxygen ( from equation)
2 moles of CH4 need 4 moles of oxygen.
1 mole of oxygen at stp occupies a volume of 22.4 litres
4 moles of oxygen at stp will have a volume of 4 x 22.4 = 89.6 litres.
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Answer:
the volume of oxygen in litres at stp required for complete combustionof 32g ch (mol.wt of c*h_{4} = 16 ) is
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