Question

Calculate the volume and amount of oxygen required for the combustion of 10 grams of methane

Answer 1

Answer:

1 mole methane(CH4)= 16 grams

10 gram= 10/16 MOLES

reaction:

CH4 +2O2 = CO2 + 2H2O

1MOLE(CH4) = 2 MOLES (O2)

10/16mole(CH4)=10/16×2 moles 02

vol=22.4 ×10/8

=22.4×1.25

=28 litres

Explanation:

at STP 1 mole = 22.414 litres