Question #284f8

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Question #284f8

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Answer 1 · 2017-09-20T17:57:10

4 moles of Aluminum

Explanation:

If we assume that we are creating Aluminum Chloride from the
metal-acid reaction between Aluminum and hydrochloric acid we can begin with the balanced chemical reaction below.

$2 A l + 6 H C l \to 2 A l C l_{3} + 3 H_{2}$ $2Al + 6HCl -> 2AlCl_3 + 3H_2$

We can use the mole ratio $\frac{2 m o l A l}{2 m o l A l C l_{3}}$ $(2 mol Al)/(2 mol AlCl_3)$

The stoichiometry equation becomes

$4 m o l A l C l_{3} \frac{2 m o l A l}{2 m o l A l C l_{3}}$ $4 mol AlCl_3 (2 mol Al)/(2 mol AlCl_3)$

$4 cancel(mol AlCl_3) (cancel2 mol Al)/(cancel2 cancel(mol AlCl_3))$

$= 4 mol Al$

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