Question

What is the percent yield for NO in a reaction with 4 moles of O2? The actual mass obtained from the reaction is 90 grams.

Answer 1

Assuming nitrogen in excess,

$N_2+O_2\to 2NO$

One mole of $O_2$ produce $2$ moles of $NO$ .So,$4$ moles of oxygen will produce $8$ moles of $NO$ .

Theoretical yield of $NO=8\times (14+16)=240\ g$

Experimental yield $=90\ g$

Percent yield$=\frac{90}{240}\times 100=37.5$ %