Respuesta :
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
From the stoichiometry of the reaction; we have at the end of the reaction, 49.2 g of AlCl3, 30.8 g of N2, 39.6 g of water and 11.55 g of Al(NO2)3.
The equation of the reaction is;
Al(NO2)3(aq) + 3 NH4Cl(aq) → AlCl3(aq) + 3 N2(g) + 6 H2O
Number of moles of Al(NO2)3 = 72.5 g/165 g/mol = 0.44 moles
Number of moles of NH4Cl = 58.6 g/53 g/mol = 1.1 moles
Since 1 mole of Al(NO2)3 reacts with 3 moles of NH4Cl
0.44 moles of Al(NO2)3 reacts with 0.44 moles × 3 moles/1 mole = 1.32 moles
Hence NH4Cl is the limiting reactant.
3 moles of NH4Cl yields 1 mole of AlCl3
1.1 moles of NH4Cl yields 1.1 × 1 mole/3 moles = 0.37 moles
Mass of AlCl3 = 0.37 moles × 133 g/mol = 49.2 g
3 moles of NH4Cl yields 3 moles of N2
1.1 moles of NH4Cl yields 1.1 moles × 3 moles/3 moles = 1.1 moles of N2
Mass of N2 = 1.1 moles of N2 × 28 g/mol = 30.8 g
3 moles of NH4Cl yields 6 moles of H2O
1.1 moles of NH4Cl yields 1.1 moles × 6 moles/3 moles = 2.2 moles of H2O
Mass of water = 2.2 moles of H2O × 18 g/mol = 39.6 g
If 3 moles of NH4Cl reacts with 1 mole of Al(NO2)3
1.1 moles of NH4Cl reacts with 1.1 moles × 1 mole /3 moles = 0.37 moles
Number of moles of excess reactant = 0.44 moles - 0.37 moles = 0.07 moles
Mass of excess reactant = 0.07 moles × 165 g/mol = 11.55 g
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