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

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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

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Answer 1 · 2014-05-05T20:32:46

You begin with the balanced chemical equation

$3 M g C l_{2} + 2 A l \to 3 M g + 2 A l C l_{3}$ $3MgCl_2 + 2Al -> 3Mg + 2AlCl_3$

The mole ratios for this equation are

3 moles $M g C l_{2}$ $MgCl_2$ : 2 moles Al : 3 moles Mg : 2 moles $A l C l_{3}$ $AlCl_3$

Therefore if 8.0 moles of $M g C l_{2}$ $MgCl_2$ react we can use the mole ration between $M g C l_{2}$ $MgCl_2$ and $A l C l_{3}$ $AlCl_3$ to determine the amount of ammonium chloride can theoretically be produced.

8.0 mol $M g C l_{2}$ $MgCl_2$ x $\frac{2 m o l A l C l_{3}}{3 m o l M g C l_{2}}$ $(2 mol AlCl_3) / (3mol MgCl_2)$ = 5.3 mol $A l C l_{3}$ $AlCl_3$

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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

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