Answer : The theoretical yield in grams for this reaction under the given conditions is, 8.28 grams.
Explanation : Given,
Mass of [tex]H_2[/tex] = 1.46 g
Mass of [tex]N_2[/tex] = 10.5 g
Molar mass of [tex]H_2[/tex] = 2 g/mol
Molar mass of [tex]N_2[/tex] = 28 g/mol
First we have to calculate the moles of [tex]H_2[/tex] and [tex]N_2[/tex].
[tex]\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}[/tex]
[tex]\text{Moles of }H_2=\frac{1.46g}{2g/mol}=0.73mol[/tex]
and,
[tex]\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}[/tex]
[tex]\text{Moles of }N_2=\frac{10.5g}{28g/mol}=0.375mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]3H_2+N_2\rightarrow 2NH_3[/tex]
From the balanced reaction we conclude that
As, 3 mole of [tex]H_2[/tex] react with 1 mole of [tex]N_2[/tex]
So, 0.73 moles of [tex]H_2[/tex] react with [tex]\frac{0.73}{3}=0.243[/tex] moles of [tex]N_2[/tex]
From this we conclude that, [tex]N_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]H_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]NH_3[/tex]
From the reaction, we conclude that
As, 3 mole of [tex]H_2[/tex] react to give 2 mole of [tex]NH_3[/tex]
So, 0.73 mole of [tex]H_2[/tex] react to give [tex]\frac{2}{3}\times 0.73=0.487[/tex] mole of [tex]NH_3[/tex]
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]
Molar mass of [tex]NH_3[/tex] = 17 g/mole
[tex]\text{ Mass of }NH_3=(0.487moles)\times (17g/mole)=8.28g[/tex]
Therefore, the theoretical yield in grams for this reaction under the given conditions is, 8.28 grams.
