Question

A `20*g` mass of oxygen gas reacts with excess `NO(g)`. How much `NO_2(g)` can be formed?

Answer 1

Approx. `60*g`.

Explanation:

`NO(g) + 1/2O_2(g) rarr NO_2(g)`

There are `(20*g)/(32.00*g*mol^-1)` `O_2(g)` `=` `0.625*mol`

Given the stoichiometry, with excess `NO`, `1.25*mol` `NO_2` can be formed.

So the mass of nitric oxide that can be produced is:

`1.25*molxx46.01*g*mol^-1` `=` `??`