Respuesta :
Answer: The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For aluminium nitrite:
Given mass of aluminium nitrite = 27.4 g
Molar mass of aluminium nitrite = 41 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol[/tex]
- For ammonium chloride:
Given mass of ammonium chloride = 169.9 g
Molar mass of ammonium chloride = 53.5 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol[/tex]
The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:
[tex]Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride
So, 0.668 moles of aluminium nitrite will react with = [tex]\frac{3}{1}\times 0.668=2.004mol[/tex] of ammonium chloride.
As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.
Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.
Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles
Calculating the mass of ammonium chloride by using equation 1, we get:
Excess moles of ammonium chloride = 1.172 moles
Molar mass of ammonium chloride = 53.5 g/mol
Putting values in equation 1, we get:
[tex]1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g[/tex]
Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams
