# If 8 moles of magnesium chloride react with enough aluminum, how many moles of aluminum chloride are produced? 3 MgCl2 + 2 Al → 3 Mg + 2 AlCl3

May 5, 2014

You begin with the balanced chemical equation

$3 M g C {l}_{2} + 2 A l \to 3 M g + 2 A l C {l}_{3}$

The mole ratios for this equation are

3 moles $M g C {l}_{2}$ : 2 moles Al : 3 moles Mg : 2 moles $A l C {l}_{3}$

Therefore if 8.0 moles of $M g C {l}_{2}$ react we can use the mole ration between $M g C {l}_{2}$ and $A l C {l}_{3}$ to determine the amount of ammonium chloride can theoretically be produced.

8.0 mol $M g C {l}_{2}$ x $\frac{2 m o l A l C {l}_{3}}{3 m o l M g C {l}_{2}}$ = 5.3 mol $A l C {l}_{3}$