Answer:
The answer to your question is
Al(NO₂)₃ = 23.08 g; NH₄Cl = 0; AlCl₃ = 71,62; N₂ = 45.06 g; H₂O = 57.94 g
Explanation:
Data
Al(NO₂)₃ = 111.6 g
NH₄Cl = 86.1 g
AlCl₃ = ?
N₂ = ?
H₂O = ?
Balanced chemical reaction
Al(NO₂)₃ + 3NH₄Cl ⇒ AlCl₃ + 3N₂ + 6H₂O
Process
1.- Calculate the molar mass of the reactants and products
Al(NO₂)₃ = 27 + (14 x 3) + (16 x 6) = 165 g
3NH₄Cl = 3[14 + 35.5 + 4] = 160.5 g
AlCl₃ = 27 + (35.5 x 3) = 133.5 g
3N₂ = 3(14 x 2) = 84 g
6H₂O = 6(2 + 16) = 108 g
2.- Calculate the limiting reactant
theoretical yield = 165/160.5 = 1.03
experimental yield = 111.6/86.1 = 1.3
From this information we conclude that the limiting reactant is NH₄Cl because the experimental yield was higher.
3.- Calculate the mass of the products
160.5 g of NH₄Cl ---------------- 133.5 g of AlCl₃
86.1 g of NH₄Cl --------------- x
x = (86.1 x 133.5)/160.5
x = 71.62 g of AlCl₃
160.5 g of NH₄Cl ---------------- 84 g of N₂
86.1 g of NH₄Cl --------------- x
x = (86.1 x 84)/160.5
x = 45.06 g of AlCl₃
160.5 g of NH₄Cl ---------------- 108 g of H₂O
86.1 g of NH₄Cl --------------- x
x = (86.1 x 108)/160.5
x = 57.94 g of AlCl₃
160.5 g of NH₄Cl ---------------- 165 g of Al(NO₂)₃
86.1 g of NH₄Cl --------------- x
x = (86.1 x 165)/160.5
x = 88.51 g of AlCl₃
Mass of Al(NO₂)₃ = 111.6 - 88.51
= 23.08 g