Respuesta :
Answer : The formula of limiting reagent is, [tex]Mg_3N_2[/tex].
The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.
The mass of excess reactant is, 36 grams.
Solution : Given,
Mass of [tex]Mg_3N_2[/tex] = 101 g
Mass of [tex]H_2O[/tex] = 144 g
Molar mass of [tex]Mg_3N_2[/tex] = 101 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
Molar mass of [tex]Mg(OH)_2[/tex] = 58 g/mole
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{ Mass of }Mg_3N_2}{\text{ Molar mass of }Mg_3N_2}=\frac{101g}{101g/mole}=1moles[/tex]
[tex]\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{144g}{18g/mole}=8moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]Mg_3N_2+6H_2O\rightarrow 3Mg(OH)_2+2NH_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Mg_3N_2[/tex] react with 6 mole of [tex]H_2O[/tex]
So, given 1 moles of [tex]Mg_3N_2[/tex] react with 6 moles of [tex]H_2O[/tex]
From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Mg_3N_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Mg(OH)_2[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]Mg_3N_2[/tex] react to give 3 mole of [tex]Mg(OH)_2[/tex]
So, given 1 mole of [tex]Mg_3N_2[/tex] react to give 3 moles of [tex]Mg(OH)_2[/tex]
Now we have to calculate the mass of [tex]Mg(OH)_2[/tex]
[tex]\text{ Mass of }Mg(OH)_2=\text{ Moles of }Mg(OH)_2\times \text{ Molar mass of }Mg(OH)_2[/tex]
[tex]\text{ Mass of }Mg(OH)_2=(3moles)\times (58g/mole)=174g[/tex]
The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.
Now we have to calculate the moles of excess reactant [tex](H_2O)[/tex].
Moles of excess reactant = 8 - 6 = 2 moles
Now we have to calculate the mass of excess reactant.
[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]
[tex]\text{ Mass of }H_2O=(2moles)\times (18g/mole)=36g[/tex]
The mass of excess reactant is, 36 grams.
