Aluminum nitrate and ammonium chloride react to form alluminum chloride, nitrogen, and water. what mass of aluminum chloride is present after 43.0 g of aluminum nitrate and 43.0 g ammonium chloride have reacted completely?
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To calculate the mass of aluminum chloride formed, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
Let's find the moles of each reactant:
1. Aluminum Nitrate (Al(NO3)3):
- Atomic mass of Aluminum (Al) = 26.98 g/mol
- Atomic mass of Nitrogen (N) = 14.01 g/mol
- Atomic mass of Oxygen (O) = 16.00 g/mol
So, the molar mass of Aluminum Nitrate (Al(NO3)3) is:
26.98 + (3 × (14.01 + 3 × 16.00)) = 213.00 g/mol
To convert mass to moles, use the formula:
moles = mass / molar mass
moles of Aluminum Nitrate = 43.0 g / 213.00 g/mol
2. Ammonium Chloride (NH4Cl):
- Atomic mass of Nitrogen (N) = 14.01 g/mol
- Atomic mass of Hydrogen (H) = 1.01 g/mol
- Atomic mass of Chlorine (Cl) = 35.45 g/mol
So, the molar mass of Ammonium Chloride (NH4Cl) is:
14.01 + (4 × 1.01) + 35.45 = 53.49 g/mol
moles of Ammonium Chloride = 43.0 g / 53.49 g/mol
Now, let's calculate the stoichiometric ratios between the reactants and the product:
The balanced equation for the reaction is:
2Al(NO3)3 + 6NH4Cl -> 2AlCl3 + 6NH3 + 9H2O + N2
So, the ratio of Aluminum Nitrate to Aluminum Chloride is 2:2 or 1:1.
From the moles of Aluminum Nitrate, we can determine the moles of Aluminum Chloride (since the ratio is 1:1):
moles of Aluminum Chloride = moles of Aluminum Nitrate
Now, let's convert the moles of Aluminum Chloride back to mass:
mass of Aluminum Chloride = moles of Aluminum Chloride × molar mass of Aluminum Chloride
The molar mass of Aluminum Chloride (AlCl3) is:
26.98 + (3 × 35.45) = 133.34 g/mol
mass of Aluminum Chloride = moles of Aluminum Chloride × 133.34 g/mol
Now, substitute the values into the equation:
mass of Aluminum Chloride = (43.0 g / 213.00 g/mol) × 133.34 g/mol
Calculate the mass of Aluminum Chloride to get the final answer.
To determine the mass of aluminum chloride formed, we need to calculate the molar mass of all the substances involved and use stoichiometry.
1. Calculate the molar mass of each compound:
- Aluminum nitrate (Al(NO3)3) = 27.0 g/mol (Al) + 3(14.0 g/mol (N) + 16.0 g/mol (O)) = 213.0 g/mol
- Ammonium chloride (NH4Cl) = 14.0 g/mol (N) + 4(1.0 g/mol (H)) + 35.5 g/mol (Cl) = 53.5 g/mol
- Aluminum chloride (AlCl3) = 27.0 g/mol (Al) + 3(35.5 g/mol (Cl)) = 133.5 g/mol
2. Determine the limiting reactant:
- To find the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The balanced equation is:
2Al(NO3)3 + 6NH4Cl → 2AlCl3 + 3N2 + 12H2O
- The stoichiometric ratio between aluminum nitrate and aluminum chloride is 2:2, and between ammonium chloride and aluminum chloride is 6:2. Therefore, the reacting moles of aluminum nitrate and ammonium chloride are both equal.
3. Calculate the moles of aluminum chloride formed:
- Since the moles of both reactants are equal, we can choose either aluminum nitrate or ammonium chloride to calculate the moles of aluminum chloride formed.
- Moles of aluminum chloride = 43.0 g / 213.0 g/mol = 0.2025 mol
4. Calculate the mass of aluminum chloride:
- Mass of aluminum chloride = Moles of aluminum chloride × Molar mass of aluminum chloride
- Mass of aluminum chloride = 0.2025 mol × 133.5 g/mol ≈ 27.03 g
Therefore, approximately 27.03 grams of aluminum chloride is present after 43.0 grams of aluminum nitrate and 43.0 grams of ammonium chloride have reacted completely.
The is a limiting reagent problem. We know that because BOTH reactants are given in the problem.
1. Write and balance the equation.
2a. Convert 43.0 g Al(NO3)3 to moles. moles = grams/molar mass.
2b. Convert 43.0 g NH4Cl to moles the same way.
3a. Using the coefficients in the balanced equation, convert moles Al(NO3)3 to moles AlCl3.
3b. Same procedure, convert moles NH4Cl to moles AlCl3.
3c. The moles from 3a and 3b likely will not be the same and one of them must be wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reactant producing that value is the limiting reagent.
4.Using the smaller value, convert moles of the product to grams. g = moles x molar mass.