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What mass of oxygen gas is required to burn completely `2.5 mol` of method? `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")`

Answer» Accroding to the equcation, 2 mol of `O_(2)` is required for every 1 mol of `CH_(4)`. Thus, we need `2xx2.5(=5.0) mol O_(2)`. Mass of 1 mol `O_(2)` is `32g`. Thus, mass of `5 mol O_(2)` is `5xx32 = 160 g`. Alternatively,
`? g O_(2) = 2.5 "mol" CH_(4) xx (2 "mol" O_(2))/(1 "mol" CH_(4)) xx (32.0g O_(2))/("1 mol" O_(2))`
`= 160 g O_(2)`


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