Respuesta :
Answer :
(a) The limiting reactant is, [tex]O_2[/tex]
(b) The mass of excess reactant is, 1.7646 g
(c) The theoretical yield of NO is, 2.616 g
(d) The percent yield of the reaction is, 37.46 %
Explanation : Given,
Mass of [tex]NH_3[/tex] = 3.25 g
Mass of [tex]O_2[/tex] = 3.50 g
Molar mass of [tex]NH_3[/tex] = 17 g/mole
Molar mass of [tex]O_2[/tex] = 32 g/mole
Molar mass of [tex]NO[/tex] = 30 g/mole
First we have to calculate the moles of [tex]NH_3[/tex] and [tex]O_2[/tex].
[tex]\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}=\frac{3.25g}{17g/mole}=0.191moles[/tex]
[tex]\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{3.50g}{32g/mole}=0.109moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
From the balanced reaction we conclude that
As, 5 moles of [tex]O_2[/tex] react with 4 mole of [tex]NH_3[/tex]
So, 0.109 moles of [tex]O_2[/tex] react with [tex]\frac{4}{5}\times 0.109=0.0872[/tex] moles of [tex]NH_3[/tex]
From this we conclude that, [tex]NH_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of product.
The limiting reactant is, [tex]O_2[/tex]
Excess moles of [tex]NH_3[/tex] = 0.191 - 0.0872 = 0.1038 mole
Now we have to calculate the mass of [tex]NH_3[/tex].
[tex]\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3[/tex]
[tex]\text{Mass of }NH_3=(0.1038mole)\times (17g/mole)=1.7646g[/tex]
The mass of excess reactant = 1.7646 g
Now we have to calculate the moles of [tex]NO[/tex].
As, 5 moles of [tex]O_2[/tex] react with 4 mole of [tex]NO[/tex]
So, 0.109 moles of [tex]O_2[/tex] react with [tex]\frac{4}{5}\times 0.109=0.0872[/tex] moles of [tex]NO[/tex]
Now we have to calculate the mass of [tex]NO[/tex].
[tex]\text{Mass of }NO=\text{Moles of }NO\times \text{Molar mass of }NO[/tex]
[tex]\text{Mass of }NO=(0.0872mole)\times (30g/mole)=2.616g[/tex]
Now we have to calculate the percent yield of [tex]NO[/tex]
[tex]\%\text{ yield of }NO=\frac{\text{Actual yield of }NO}{\text{Theoretical yield of }NO}\times 100=\frac{0.98g}{2.616g}\times 100=37.46\%[/tex]
The percent yield of the reaction is, 37.46 %
