Answer: 0.071 moles of excess reactant [tex](O_2)[/tex] will be left over.
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of [tex]NO[/tex]
[tex]\text{Number of moles}=\frac{21.6g}{30.01g/mol}=0.720moles[/tex]
b) moles of [tex]O_2[/tex]
[tex]\text{Number of moles}=\frac{13.8g}{32g/mol}=0.431moles[/tex]
[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]
According to stoichiometry :
2 moles of [tex]NO[/tex] require 1 mole of [tex]O_2[/tex]
Thus 0.720 moles of [tex]NO[/tex] require=[tex]\frac{1}{2}\times 0.720=0.360moles[/tex] of [tex]O_2[/tex]
Thus [tex]NO[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess regaent.
moles of [tex]O_2[/tex] left = (0.431-0.360) = 0.071 moles
