Respuesta :
Answer: The number of moles of excess reactant [tex]O_2[/tex] left over will be, 0.089 moles.
Explanation : Given,
Mass of [tex]NO[/tex] = 20.5 g
Mass of [tex]O_2[/tex] = 13.8 g
Molar mass of [tex]NO[/tex] = 30 g/mol
Molar mass of [tex]O_2[/tex] = 32 g/mol
First we have to calculate the moles of [tex]NO[/tex] and [tex]O_2[/tex].
[tex]\text{Moles of }NO=\frac{\text{Given mass }NO}{\text{Molar mass }NO}[/tex]
[tex]\text{Moles of }NO=\frac{20.5g}{30g/mol}=0.683mol[/tex]
and,
[tex]\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}[/tex]
[tex]\text{Moles of }O_2=\frac{13.8g}{32g/mol}=0.431mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2NO+O_2\rightarrow 2NO_2[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]NO[/tex] react with 1 mole of [tex]O_2[/tex]
So, 0.683 moles of [tex]NO[/tex] react with [tex]\frac{0.683}{2}=0.342[/tex] moles of [tex]O_2[/tex]
From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]NO[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of excess reactant [tex]O_2[/tex] left over.
Number of moles of excess reactant [tex]O_2[/tex] left over = Given moles - Required moles
Number of moles of excess reactant [tex]O_2[/tex] left over = 0.431 mol - 0.342 mol
Number of moles of excess reactant [tex]O_2[/tex] left over = 0.089 mol
Therefore, the number of moles of excess reactant [tex]O_2[/tex] left over will be, 0.089 moles.
