Question

If 5.3g O2 reacts with NH3 to produce 3.1g NO2, what would the percent yield of NO2 be?

4NH3(g) + 7O2(g) --> 4NO2 + 6H2O(g)

Answer 1

71.2%, i could be wrong.
\[5.3O2*(1mol/32g O2)*4NO2/7O2*46gNO/1mol=0.71\]

Answer 2

that gives you the grams of teorical NO2, then you divide 3.1 between the teorical grams

Answer 3

It's 71%...

Answer 4

:D so i was right hahaha

Answer 5

yes, and noticing that makes you even smarter! question closed... :-)