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:

$\begin{gathered} M_{AlCl_3}=\frac{m_{AlCl_3}}{n_{AlCl_3}} \\ n_{AlCl_3}=\frac{m_{AlCl_3}}{M_{AlCl_3}}=\frac{100g}{133.341g/mol}=0.749957\ldots mol \end{gathered}$

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

2 mol AlCl₃ --- 3 mol CaCl₂

0.749957... mol AlCl₃ --- x

$\begin{gathered} \frac{2}{0.749957\ldots mol}=\frac{3}{x} \\ 2x=2.24987\ldots mol \\ x=1.12494\ldots mol \end{gathered}$

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.