Question

Given this equation: 3 MgCl2 + 2 Al → 3 Mg + 2 AlCl3, if 2.1 moles of magnesium chloride reacted how many moles of aluminum chloride are produced?

Answer 1

Let's see that 3 moles of magnesium chloride (MgCl2) produce 2 moles of aluminum chloride (AlCl3), we can do a rule of three to find the number of moles required of AlCl3 by 2.1 moles of MgCl2, like this:

`\begin{gathered} \text{3 moles MgCl}_2\to2molesAlCl_3 \\ 2.1molesMgCl_2\to?molesAlCl_3 \end{gathered}`

The calculation would be:

`2.1molesMgCl_2\cdot(2molesAlCl_3)/(3molesMgCl_2)=1.4molesAlCl_3.`

So, the answer is that 1.4 moles of aluminum chloride are produced by 2.1 moles of magnesium chloride.