# If you began with 9.23 moles of hydrochloric acid and excess amounts of aluminum, how many grams of aluminum chloride would be formed?

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This is a stoichometry problem, in which you use the relationship between reactants and/or products in a chemical equation to calculate the quantity of one substance base on the quantity of any other substance in the reaction. These relationships are given by the coefficients in the balanced equation, so it's necessary to begin by writing and balancing the equation. The equation for this reaction is:

`2 Al + 6 HCl -gt 2 AlCl_3 + 3 H_2`

The mole ratio of AlCl3 to HCl is 2:6, so the moles of AlCl3 formed is:

9.23 moles HCl x (2 AlCl3)/(6 HCl) =  3.08 moles AlCl3

Multiplying moles of AlCl3 by its molar mass gives grams:

3.08 moles AlCl3 x 133 grams/1 mole = 410. grams AlCl3

You can solve problems like this more quickly be multiplying the given quantity by a series of conversion factors, without having to calculate the intermediate step(s):

9.23 moles HCl x (2 AlCl3)/(6 HCl) x 133g/1mol = 410. grams AlCl3

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## Related Questions

First you will need to write out your chemical equation. Combining hydrochloric acid and aluminum produces aluminum chloride and excess hydrogen.

`HCl+Al => AlCl_3 + H_2`

Next, you will need to balance the equation so there is the same amount of each substance on each side of the equation making this:

`6HCl+2Al => 2AlCl_3 + 3H_2 `

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Now we use stoichimetry and the mole ratio aluminum chloride to set up your equation like so:

`9.23mol HCl X ("2AlCl_3" / "6HCl") X("133.34gAlCl_3" / "1molAlCl_3")`

Notice how the units cancel out, leaving you with grams of aluminum chloride, the unit you are trying to find.

Simply run the math of:
` 9.238 X (2/6) X 133.34`

Your answer is 410.2427333. Round to the number of significant figures provided in the question (3) and your answer is 410 grams of aluminum chloride!