Answer to Question #298336 in General Chemistry for Katie

Question

If 10.0 moles of O₂ are reacted with excess NO in the reaction below, and only 6.4 mol of NO₂ were collected, then what is the percent yield for the reaction?

2 NO (g) + O₂ (g) → 2 NO₂ (g)

Answer 1

Moles of O2 = 10.0mol

Mole ratio of O2:NO2 = 1:2

Theoretical moles of NO2 $=\dfrac{2}{1}x10.0 mol$ = 20.0mol

Experimental moles = 6.4mol

% yield $=\dfrac{experimental}{theoretical}x100$

$=\dfrac{6.4mol}{20.0mol}x100$

= 32%