Respuesta :
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
Answer:
There remains 23.0 grams Al(NO2)3
Mass of products:
Mass AlCl3 = 71.6 grams
Mass N2 = 45.1 grams
Mass H2O = 58.0 grams
Explanation:
Step 1: Data given
Mass of aluminium nitrite= 111.6 grams
Molar mass aluminium nitrite = 165.0 g/mol
Mass of ammonium chloride = 86.1 grams
Molar mass of ammonium chloride = 53.49 g/mol
Step 2: The balanced equation
Al(NO2)3 + 3NH4Cl → AlCl3 + 3N2 + 6H2O
Step 3: Calculate moles Al(NO2)3
Moles Al(NO2)3 = mass / molar mass
Moles Al(NO2)3 = 111.6 grams / 165.0 g/mol
Moles Al(NO2)3 = 0.6764 moles
Step 4: Calculate moles NH4Cl
Moles NH4Cl = 86.1 grams / 53.49 g/mol
Moles NH4Cl = 1.61 moles
Step 5: Calculate limiting reactant
For 1 mol Al(NO2)3 we need 3 moles NH4Cl to produce 1 mol AlCl3, 3 moles N2 and 6 moles H2O
NH4Cl is the limiting reactant. IT will completely be consumed (1.61 moles).
Al(NO2)3 is in excess. There will react 1.61/3 = 0.537 moles
There will remain 0.6764 - 0.537 = 0.1394 moles
This is 0.1394 * 165.0 g/mol = 23.0 grams
Step 6: calculate moles of products
For 3 mol NH4Cl we need 1 mol Al(NO2)3 to produce 1 mol AlCl3, 3 moles N2 and 6 moles H2O
For 1.61 moles of Al(NO2)3 we'll have:
1.61/ 3 = 0.537 moles AlCl3
1*1.61 = 1.61 moles N2
2*1.61 = 3.22 moles H2O
Step 7: Calculate mass of products
Mass =moles * molar mass
Mass AlCl3 = 0.537 moles *133.34 g/mol = 71.6 grams
Mass N2 = 1.61 moles * 28.0 g/mol = 45.1 grams
Mass H2O = 9.66 moles * 18.02 g/mol = 58.0 grams
