Question

3Ca + 2AICI3 —> 3CaCI2 + 2AIIf you react 100, grams of aluminum chloride, AICI3, with excess calcium, how many grams of calcium chloride,CaCI2, are produced ?

Answer 1

Firstly, we need to convert the 100 g of AlCl₃ to number of moles, using:

`M_(AlCl_3)=\frac{m_(AlCl_3)}{n_{AlCl_(3)}}`

The molar weight of AlCl₃ is calculating consulting the atomic weights:

`\begin{gathered} M_(AlCl_3)=1\cdot M_(Al)+3\cdot M_(Cl) \\ M_(AlCl_3)=(1\cdot26.982+3\cdot35.453)g/mol \\ M_(AlCl_3)=(26.982+106.359)g/mol \\ M_(AlCl_3)=133.341g/mol \end{gathered}`

Thus:

`\begin{gathered} M_(AlCl_3)=(m_(AlCl_3))/(n_(AlCl_3)) \\ n_(AlCl_3)=(m_(AlCl_3))/(M_(AlCl_3))=(100g)/(133.341g/mol)=0.749957\ldots mol \end{gathered}`

For each 2 moles of AlCl₃, we produce 3 moles of CaCl₂. So:

2 mol AlCl₃ --- 3 mol CaCl₂

0.749957... mol AlCl₃ --- x

`\begin{gathered} (2)/(0.749957\ldots mol)=(3)/(x) \\ 2x=2.24987\ldots mol \\ x=1.12494\ldots mol \end{gathered}`

Now, we convert back to grams, but now we need the molar weight of CaCl₂:

`\begin{gathered} M_(CaCl_2)=1\cdot M_(Ca)+2\cdot M_(Cl) \\ M_(CaCl_2)=(1\cdot40.078+2\cdot35.453)g/mol \\ M_(CaCl_2)=(40.078+70.906)g/mol \\ M_(CaCl_2)=110.984g/mol \end{gathered}`

Thus:

`\begin{gathered} M_{CaCl_(2)}=\frac{m_(CaCl_2)}{n_{CaCl_(2)}} \\ m_(CaCl_2)=M_(CaCl_2)\cdot n_(CaCl_2)=110.984g/mol\cdot1.12494\ldots mol=124.850\ldots g\approx125g \end{gathered}`

So, 100 g of AlCl₃ will produce approximately 125 g of CaCl₂.