Answer :
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.
[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.
