Answer
$NO$ is the limiting reactant, and 0.886 mole of $NO_2$ is produced.
Work Step by Step
Find the amount of product if each reactant is completely consumed.
$$ 0.886 \space mole \space NO \times \frac{ 2 \space moles \ NO_2 }{ 2 \space moles \space NO } = 0.886 \space mole \space NO_2 $$
$$ 0.503 \space mole \space O_2 \times \frac{ 2 \space moles \ NO_2 }{ 1 \space mole \space O_2 } = 1.01 \space moles \space NO_2 $$
Since the reaction of $ NO $ produces less $ NO_2 $ for these quantities, it is the limiting reactant, and the reaction produces 0.886 mole of $NO_2$
