Answer:
48.95g of AlCl3
Explanation:
Balancing the equation:
2Al + 3Cl2 --> 2AlCl3
Firstly, we have to find the limiting reagent by calculating the individual moles of the reactants.
Moles = mass/molecular weight
For chlorine gas,
Molecular weight = 35.5 * 2
= 71g/mol
Moles = 39/71
= 0.55 mol
For Aluminium foil,
Molecular weight = 27 g/mol
Mole = 34/27
= 1.26mol
Comparing both Cl2 and Al moles values, chlorine gas is the limiting reagent.
By stoichiometry, since 3 moles of Cl2 will produce 2 moles of AlCl3
Moles of AlCl3 = (2/3) * 0.55
= 0.367moles
Therefore, mass of AlCl3 =
Moles * molecular weight
Molecular weight = 27 + (3 * 35.5)
= 133.5 g/mol
Mass = 0.367 * 133.5
= 48.95g of AlCl3
