Number of moles formed: How many moles of water are produced when `4.5m` mol of methance reacts with excess oxygen?
Strategy: According to the balanced equation for the combusion of methane,
`CH_(4) + 2O_(2) rarr CO_(2) + 2H_(O)`
1 mol of methance alwyas reacts with 2 mol of oxygen to produce 2 mol of water. Either we use unitary approach or factor- we use unitary approah or factor-label method by constructing the unit factor.
`(2 mol H_(2) IO)/(1 mol CH_(4))`

Explanation

Unitary method
`1 mol CH_(4)` produces `2 mol H_(2) O`
`:. 4.5 mol CH_(4)` will produce `2(4.5) = 9.0 mol H_(2) O`
Factor-label method
`?mol H_(2) O = 4.5 mol CH_(4) xx (2 mol H_(2) O)/(1 mol CH_(4))`
`= 9.0 mol H_(2)O`
Because moles can be converted to mass, we can also gives a mass interpection of a chemical equation. We known the mass of 1 mol of each of the substances in the combusion reaction. So, we can also write
`CH_(4) + 2O_(2) rarr CO_(2) + 2H_(0) O`
`16.0 g 2xx32g 44.0g 2xx180.0g` (mass interpection)
`ubrace(CH_(4)+ O_(2))_("80.0g reactants") ubrace(CO_(2)+ H_(2)O)_("80.0g products")`
The equation now tells us that `16.0g` of `CH_(4)` reacts with `64.0g` of `O_(2)` to from `44.0g` of `CO_(2)` and `36.0g` of `H_(2) O`. Note that a balance equation satisfies the law of conservation of mass.