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.