Answer :
Answer : The mass of nitrogen dioxide is, 17.6 grams
The formula of limiting reagent is, [tex]NO[/tex]
The mass of excess reagent remains is, 3.74 grams
Explanation : Given,
Mass of [tex]NO[/tex] = 11.5 g
Mass of [tex]O_2[/tex] = 9.91 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{11.5g}{30g/mol}=0.383mol[/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{9.91g}{32g/mol}=0.309mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/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, 0.383 moles of [tex]NO[/tex] react with [tex]\frac{0.383}{2}=0.192[/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.
Moles of excess reagent remains = 0.309 - 0.192 = 0.117 mol
Now we have to calculate the mass of excess reagent remains.
[tex]\text{ Mass of excess reagent}=\text{ Moles of excess reagent}\times \text{ Molar mass of excess reagent}[/tex]
Molar mass of [tex]O_2[/tex] = 32 g/mole
[tex]\text{ Mass of excess reagent}=(0.117moles)\times (32g/mole)=3.74g[/tex]
Now we have to calculate the moles of [tex]NO_2[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]NO[/tex] react to give 2 mole of [tex]NO_2[/tex]
So, 0.383 mole of [tex]NO[/tex] react to give 0.383 mole of [tex]NO_2[/tex]
Now we have to calculate the mass of [tex]NO_2[/tex]
[tex]\text{ Mass of }NO_2=\text{ Moles of }NO_2\times \text{ Molar mass of }NO_2[/tex]
Molar mass of [tex]NO_2[/tex] = 46 g/mole
[tex]\text{ Mass of }NO_2=(0.383moles)\times (46g/mole)=17.6g[/tex]