Question

A `21.3*g` mass of `"nitrous oxide"`, `N_2O`, is oxidized by dioxygen gas to `NO_2(g)`. How much dioxygen is required?

Answer 1

`N_2O(g) + 3/2O_2(g)rarr2NO_2(g)`

Explanation:

`"Moles of"` `N_2O` `=` `(21.3*g)/(44.02*g*mol^-1)=0.483*mol`

And for stoichiometric reaction we need,

`0.483*molxx3/2xx32.00*g*mol^-1=??*g` `"dioxygen gas"`.