Question

How many grams of Cl2 must react to produce 0.0923 mol of AlCl3? Show work.

Answer 1

First, let's state the chemical reaction:

$2Al+3Cl_2\to2AlCl_3\text{.}$

We can find the number of moles of Cl2 required to produce 0.0923 moles of AlCl3, doing a rule of three: 3 moles of Cl2 reacted produces 2 moles of AlCl3:

$\begin{gathered} 3molesCl_2\to2molesAlCl_3 \\ \text{?moles Cl}_2\to0.0923\text{ moles }AlCl_3\text{.} \end{gathered}$

The calculation would be:

$0.0923molesAlCl_3\cdot\frac{3molesCl_2}{2molesAlCl_3}=0.138molesCl_2.$

And the final step is to convert this number of moles to grams. Remember that the molar mass can be calculated using the periodic table, so the molar mass of Cl2 is 70.8 g/mol, and the conversion is:

$0.138molesCl_2\cdot\frac{70.8gCl_2}{1molCl_2}=9.770gCl_2.$

The answer is that we need 9.770 grams of Cl2 to produce 0.0923 moles of AlCl3.