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How many moles of [tex]KCl[/tex] are produced at the same time as 9 moles of [tex]O_2[/tex]?

[tex]\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \][/tex]

[tex]\[ \text{[?] mol KCl} \][/tex]


Sagot :

To determine how many moles of [tex]\( KCl \)[/tex] are produced when 9 moles of [tex]\( O_2 \)[/tex] are generated, we will use the given balanced chemical equation:

[tex]\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \][/tex]

This equation tells us the stoichiometric relationship between [tex]\( KClO_3 \)[/tex], [tex]\( KCl \)[/tex], and [tex]\( O_2 \)[/tex]. In particular, for every 2 moles of [tex]\( KClO_3 \)[/tex] decomposed, 2 moles of [tex]\( KCl \)[/tex] and 3 moles of [tex]\( O_2 \)[/tex] are produced.

First, let’s identify the mole ratio between [tex]\( KCl \)[/tex] and [tex]\( O_2 \)[/tex]:

From the equation, we see that 3 moles of [tex]\( O_2 \)[/tex] are produced for every 2 moles of [tex]\( KCl \)[/tex]. Mathematically, we can write this ratio as:

[tex]\[ \frac{2 \text{ moles } KCl}{3 \text{ moles } O_2} \][/tex]

Now we know that we have 9 moles of [tex]\( O_2 \)[/tex], what we need to do is use the stoichiometric ratio to find how many moles of [tex]\( KCl \)[/tex] will be produced from these 9 moles of [tex]\( O_2 \)[/tex].

To determine the number of moles of [tex]\( KCl \)[/tex], we can set up the following proportion using the ratio we identified earlier:

[tex]\[ \frac{2 \text{ moles } KCl}{3 \text{ moles } O_2} = \frac{x \text{ moles } KCl}{9 \text{ moles } O_2} \][/tex]

Solving for [tex]\( x \)[/tex], we have:

[tex]\[ x = \left( \frac{2 \text{ moles } KCl}{3 \text{ moles } O_2} \right) \times 9 \text{ moles } O_2 \][/tex]

[tex]\[ x = \frac{2}{3} \times 9 \][/tex]

[tex]\[ x = 2 \times 3 \][/tex]

[tex]\[ x = 6 \][/tex]

Therefore, 6 moles of [tex]\( KCl \)[/tex] are produced when 9 moles of [tex]\( O_2 \)[/tex] are generated.
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