Explanation:
Given Data:
The mass of aluminium nitrite is 72.5 g
The mass of ammonium chloride is 58.6 g
The balanced chemical equation for the reaction is given as follows.
Al(NO2)3 + 3NH4Cl → AlCl3 + 3N2 + 6H2O
The number of moles can be determined by the formula given as follows.
Number of moles = Mass / Molar mass
The molar mass of aluminum nitrate and ammonium chloride is 164.998 g/mol and 53.49 g/mol respectively.
inserting the respective values in the formula given above.
Moles of Al(NO2)3 = 72.5 g / 164.998 g/mol = 0.439 mol
Moles of NH4Cl = 58.6 g / 53.49 g/mol = 1.096 mol
From the balanced equation,
3 moles of ammonium chloride requires 1 mole of aluminum nitrate.
So, 1 mole of ammonium chloride requires 1 / 3 mole of aluminum nitrate.
Thus, 1.096 mole of ammonium chloride will require (1 / 3) × 1.096 = 0.3653 mole of aluminum nitrite.
Here, the amount of aluminum nitrate is more than the required amount so ammonium chloride is the limiting reagent.
From the balanced chemical equation, 3 mole of ammonium chloride gives 1 mole of aluminum chloride.
So, 1.096 mole of ammonium chloride will give;
3 = 1
1.096 = x
x = (1.096 * 1 ) / 3 = 0.3653 mole of aluminum chloride.
Therefore, the number of moles of aluminum chloride is 0.3653 mol.
Since the molar mass of aluminum chloride is 133.34 g/mol
Substitute the respective values in the formula given above.
0.3653 mol = Mass / 133.34 g/mol
Mass = 0.3653 mol × 133.34 g/mol = 48.71 g
Therefore , the mass of aluminum chloride produces is 48.71 g.
The Ammonium chloride is completely used up in the reaction.
The amount of alumium nitrite used is = Number of moles * Molar mass = 0.3653 * 164.998 = 60.27
Mass of alminium nitrite left = 72.5 - 60.27 = 12.23g
