What mass of oxygen gas is required to burn completely `2.5 mol` of methane?
`underset(underset("reactant")("one"))("Moles of ")rarr underset(underset("reactant")("another"))("Mass of")`
Stragetgy : Use the balanced equation
`{:(CH_(4), + 2O_(2), rarr CO_(2), + 2H_(2) O),(1mol, 2mol,1 mol,2mol),(16.0 g,2(32.0)g,44.0g,2(18.0g)):}`
to find the relationship among moles and grams of reactants :
`underset(CH_(4))("Mole of ")rarr underset(O_(2))("Moles of")rarr underset(O_(2))("grams of")`

Explanation

We can solve the problem in three steps.
Step1: Mass `CH_(4) rarr mol CH_(4)`
`? "mol" CH_(4) = 40.0g CH_(4) xx (1 mol CH_(4))/(16.0g CH_(4)) (n = ("Mass")/("Molar mass"))`
`= 2.50 "mol" CH_(4)`
Step 2 : `"mol" CH_(4) rarr mol O_(2)`
`? "mol" O_(2) = 2.50 "mol" CH_(4) xx (2 "mol" O_(2))/(1 "mol" CH_(4))`
(Balanced equation used)
`= 5.00 "mol" O_(2)`
Step 3 : mol `O_(2) rarr` mass `O_(2)`
`? g O_(2) = 5.00 "mol" O_(2) xx (32.0 g O_(2))/(1 mol O_(2)) ("Mass = n xx MM")`
`= 160g O_(2)`
Alternaty all these steps can be combined into one setup as follows :
`underset(CH_(4))("g of")rarrunderset(CH_(4))("mol of")rarr underset(O_(2))("mol of")rarr underset(O_(2))("g of")`
`? g O_(2) = 40.0g CH_(4) xx (1 "mol" CH_(4))/(16.0g CH_(4)) xx (2 "mol" O_(2))/(1 "mol" CH_(4))`
`xx (32.0 g O_(2))/("1 mol" O_(2))`
`= 160g O_(2)`