ow many grams of aluminum chloride must decompose to produce 78.3 milliliters of aluminum metal, if the density of aluminum is 2.70 g/mL? Show all steps of your calculation as well as the final answer. (5 points)
AlCl3 → Al + Cl2
2AlCl3 → 2Al + 3Cl2
mass = volume x density
mass = 78.3 x 2.70 = approx 211 g but you need a more accurate answer than my estimate.
mols Al = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Al to mols AlCl3.
Then g AlCl3 = mols AlCl3 x molar mass AlCl3
can you do this with the numbers please?
To determine the amount of aluminum chloride that must decompose to produce 78.3 milliliters of aluminum metal, we can use the given density of aluminum and the molar ratio in the balanced chemical equation.
1. Determine the molar mass of aluminum chloride (AlCl3):
- The atomic mass of aluminum (Al) is 26.98 g/mol.
- The atomic mass of chlorine (Cl) is 35.45 g/mol.
- Since aluminum chloride (AlCl3) contains 1 aluminum atom and 3 chlorine atoms:
Molar mass of AlCl3 = (1 * 26.98 g/mol) + (3 * 35.45 g/mol) = 133.34 g/mol.
2. Convert the given volume of aluminum metal (78.3 mL) into grams:
- Given density of aluminum = 2.70 g/mL.
- Mass of aluminum = volume of aluminum metal * density of aluminum
= 78.3 mL * 2.70 g/mL = 211.41 g.
3. Determine the molar mass of aluminum:
- The atomic mass of aluminum (Al) is 26.98 g/mol.
4. Find the molar ratio between aluminum chloride and aluminum:
- Balanced chemical equation: AlCl3 → Al + Cl2.
- The stoichiometry shows that 1 mole of AlCl3 produces 1 mole of Al.
5. Calculate the moles of aluminum:
- Moles of Al = mass of Al / molar mass of Al
= 211.41 g / 26.98 g/mol = 7.8372 mol.
6. Use the molar ratio to find the moles of aluminum chloride:
- Moles of AlCl3 = Moles of Al.
7. Calculate the mass of aluminum chloride:
- Mass of AlCl3 = Moles of AlCl3 * molar mass of AlCl3
= 7.8372 mol * 133.34 g/mol = 1045.82 g.
Therefore, 1045.82 grams of aluminum chloride must decompose to produce 78.3 milliliters of aluminum metal.