Answer to Question #137674 in Chemistry for Carl Shem Lavalle

Question

What is the mass of AlCl3 if there are 34.0 g of aluminum that is reacted to 39.0 g of chlorine gas?

Answer 1

The balanced equation for the reaction between aluminum and chlorine is written as follows:

2Al + 3Cl₂ $\rightarrow$ 2AlCl₃.

From this equation, one can see that 2 moles of aluminum react with 3 moles of chlorine gas and 2 moles of aluminum chloride is formed.

Let's calculate the number of the moles of aluminum and of chlorine gas available for reaction:

$n(Al) = \frac{m}{M} = \frac{34 \text{ g}}{26.98 \text{ g/mol}} = 1.26$ mol, and

$n(Cl_2) = \frac{m}{M} = \frac{39 \text{ g}}{70.91 \text{ g/mol}} =0.550$ mol.

If we analyze the stoichiometry of the reaction, we must compare the number of the moles of aluminum and that of chlorine in the following way:

$\frac{n(Al)}{2} = 0.630$

$\frac{n(Cl_2)}{3} = 0.183$ .

As one can see from these relations, 0.630>0.183, so aluminum is in excess. Therefore, the number of the moles of AlCl₃ produced must be calculated from the number of the moles of chlorine available for reaction:

$n(AlCl_3) = 2·\frac{n(Cl_2)}{3} = 0.367$ mol.

Finally, the mass of AlCl₃ ($M = 133.34$ g/mol) produced is:

$m(AlCl_3) = nM = 0.367\text{ mol}·133.34\text{ g/mol}$

$m(AlCl_3) = 48.9$ g.

Answer: if there are 34.0 g of aluminum that is reacted to 39.0 g of chlorine gas, the mass of AlCl₃ is 48.9 g.