Methane undergoes slow atmospheric oxidation and produces carbonmonoxide. If `2xx10^(22)` oxygen molecules are used in such oxidation, what weight of methane is consumed?

Explanation

Slow atmospheric oxidation of methane is given by the equation
`2CH_(4)+3O_(2) to 2CO+4H_(2)O`
3 moles of `O_(2)="2 moles of "CH_(4)`
`3 xx 6.022 xx 10^(23) O_(2)` molecules =
The weight of `CH_(4)` consumed `(2 xx 16)"grams of "CH_(4)`
`2 xx 10^(22) O_(2)` molecules=?
The weight of `CH_(4)` consumed `=(2 xx 10^(22) xx 32)/(3 xx 6.022 xx 10^(23))=0.354 g`