Respuesta :
Answer : The correct option is, 23.6 g
Explanation : Given,
Mass of [tex]Mg_3N_2[/tex] = 100.0 g
Mass of [tex]H_2O[/tex] = 75.0 g
Molar mass of [tex]Mg_3N_2[/tex] = 101 g/mol
Molar mass of [tex]H_2O[/tex] = 18 g/mol
First we have to calculate the moles of [tex]Mg_3N_2[/tex] and [tex]H_2O[/tex].
[tex]\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}[/tex]
[tex]\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol[/tex]
and,
[tex]\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}[/tex]
[tex]\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)[/tex]
From the balanced reaction we conclude that
As, 6 moles of [tex]H_2O[/tex] react with 1 mole of [tex]Mg_3N_2[/tex]
So, 4.17 moles of [tex]H_2O[/tex] react with [tex]\frac{4.17}{6}=0.695[/tex] moles of [tex]Mg_3N_2[/tex]
From this we conclude that, [tex]Mg_3N_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]H_2O[/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, 6 moles of [tex]H_2O[/tex] react to give 2 moles of [tex]NH_3[/tex]
So, 4.17 moles of [tex]H_2O[/tex] react to give [tex]\frac{2}{6}\times 4.17=1.39[/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=(1.39moles)\times (17g/mole)=23.6g[/tex]
Therefore, the maximum theoretical yield of [tex]NH_3[/tex] is, 23.6 grams.
