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

1 Answer
Apr 20, 2016

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 = ??