luminum reacts with chlorine gas to form aluminum chloride via the following reaction: Al + Cl2 → AlCl3 How many grams of aluminum chloride could be produced from 34.0 g of aluminum and 39.0 g of chlorine gas?

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