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In the reaction:

$$\ce{Al_2(SO_4)_3 + 3Ca(OH)_2 -> 2Al(OH)_3 + 3CaSO_4}$$

How many moles of $\ce{Al_2(SO_4)_3}$ react with $6~\mathrm{mol}$ of $\ce{Ca(OH)_2}$?

I want to say $2$, but I am unsure. If one mole of $\ce{Al_2(SO_4)_3}$ reacts with $3~\mathrm{mol}$ of $\ce{Ca(OH)_2}$, then $\frac{6}{3} = 2$, but I just don't know if this is correct.

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Exactly, why wouldn't it be? You could write it as a proportion: 1 : 3 = x : 6

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    $\begingroup$ I'm not sure, I'm just afraid I'll think incorrectly about a question. I am new to stoichiometry. $\endgroup$ – None Aug 29 '14 at 10:37
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Here're my assumptions. I believe you have a some basic understanding of cations, anions, valence electrons, polarity and how covalent/ionic/hydrogen bonds are made. So I assumed you know how to write an accurate chemical reaction as an equation by balancing left side with right side. I will do my best to write this in laymen as well as some form of a chemical language. I also suggest you get your basics of ratios polished.

Mole is a reference unit to a count like,

a dozen (of cup cakes) refers to 12
a couple (of minutes) refers to 2
a pair (of shoes) refers to 2

In chemistry the mole has the same concept behind. Mole of any element has $6.022 \times10^{23}$ of that element's atoms. Same goes for a compound. Let's take salt. I took salt ($NaCl$), because it's rather a simple compound compared to sugar.

Let's say you go for shopping, and ask the grocer "hello there, can I have a mole of salt?" What are you expecting here? How many molecules of NaCl is in 1 mole? $6.022 \times10^{23}$ NaCl molecules. If the grocer had some understanding of chemistry, he may have said "alright" but then he may have to use a table spoon to measure, given most grocers do not have a chemical scale around.

Now if I ask you, how many Sodiumn Na atoms are in 1 mole of Na? That's $6.022 \times10^{23}$ of Na atoms. If we look at the formula for making NaCl, 1 mole of Na (reactant) and 1 mole of Cl (reactant) makes 1 mole of salt (product).

$\ce{1\,Na} + \ce{1\,Cl} = \ce{1\,NaCl}$

Always remember the Empirical Formula, the base unit (my own reference) of a reaction. What I meant by base unit is the the atomic ratio among the elements and final compound as shown in the following equation for salt. When you know the base unit's atomic ratio then you can simply calculate to get the number of reactants, products in the reaction.(higher, lower proportions).

If we take the atomatic ratios of reactants and product.

1 Na to 1 Cl, can be written as 1 Na : 1 Cl = 1 : 1 1 Na to 1 NaCl 1 Cl to 1 NaCl and so on.

So if you want 2 moles of salt, you must have 2 moles of Na and 2 moles of Cl respectively.

All you have to do is multiply base unit of the equation by 2 here,

$ 2\times (1\,Na + 1\,Cl = 1\,NaCl)$

Resulting,

$2\,Na + 2\,Cl = 2\,NaCl$

Similarly, let's take $H_2O$ (water). One mole of Oxygen, 2 moles of Hydrogen make water.

$2\,H + 1\,O = 1\,H_2O$

Infact 1 mole of water is not enough to quench your thirst :). So let's say you want to make half mole of water. How would you go about it? Above is the base unit of making water. And you multiply the entire equation by 0.5 or 1/2.

$ (0.5)\times (2\,H + 1\,O = 1\,H_2O)$

resulting,

$ 1\,H + 0.5\,O = 0.5\,H_2O$

If I ask you, how many Hydrogen moles are reacting with 10 moles of Oxygen. What will you do? Go back to the base unit and to the atomic ratio of the reactants. Remember any new amount of moles that you want to use for an element in this reaction, it's proportional to the base unit's atomic ratio of the reactants and products.

1 Oxygen react with 2 Hydrogen. 2 Hydrogen to 1 Oxygen = ?Hydrogen to 10 Oxygen $2\,H : 1\,O = ?\,H : 10\,O$

By writing it as a fraction,

$\frac{2\,H}{1\,O} = \frac{?\,H}{10\,O}$

$2\times 10\,H = ?\,H$

$20\,H = ?\,H$ , simply 20 Hydrogen moles.

Now let's take a look at your question.

$1\,Al_2(SO_4)^3+3\,Ca(OH)^2⟶2\,Al(OH)^3+3\,CaSO_4$

Above is the so called base unit of your chemical reaction. It is balanced and an equation. So can you question yourself,

  • how many moles of Aluminum Sulphate $Al_2(SO_4)^3$ is used in this reaction? (1 mole)
  • how many moles of Calcium Hydroxide $3Ca(OH)^2$ is used in this reaction? (3 moles)

Write down the atomic ratio of base unit for the reactants.

3 moles of $Ca(OH)^2$ to 1 mole of $Al_2(SO_4)^3 $

I am going to refer x for $Ca(OH)^2$ and y for $Al_2(SO_4)^3 $.

$ 3\,x : 1\,y = 6\,x : ?\,y $

Write down as a fraction,

$ \frac{3\,x}{1\,y} = \frac{6\,x}{?\,y} $

Are you getting there? Feel free to shoot a comment if you are still having doubts.

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