Question

Calculate the volumme of oxygen required for the complete combustion of 8.8g of propane

Answer 1

C3H8+5 O2----------------->3 CO2+ 4 H2O

No of moles of CH4=8.8/44 =0.2 moles (GM wt of propane=44 g)

1 mole of CH4------------5 moles of O2

0.2 moles----------------?

              = 1 mole* 22.4 liters =22.4 liters

Answer 2

your answer

after balancing

c3h8+ 5o2 → 3co2 +4h2o

now moles of propane

moles =mass/mW

moles=8.8/44=0.2

now

c3h8 reacts with 5 o2

moles =v/22.4

0.2×22.4=4.48