Question

What is the mass of oxygen, in grams, required to completely react a 24 gram sample of methane gas?

Answer 1

Answer: 96 g O2

Step-by-step explanation:

this is combustion

CH4 + 2 O2 --> CO2 + 2 H2O

24 g CH4 X 1 mole Ch4 / 16.043 g CH4) X ( 2 moles O2 / 1 moles CH4) X

( 31.999 G O2 / 1 mole O2 ) = 95.74 g O2 = 96 g O2 in correct sig figs