How many moles of potassium chlorate must be used to produce 6 moles of oxygen gas?

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How many moles of potassium chlorate must be used to produce 6 moles of oxygen gas?

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Answer 1 · 2016-12-21T04:03:03

Following the equation:
$2 K C l O_{3} \Rightarrow 2 K C l + 3 O_{2}$ $2KClO_3 => 2KCl + 3O_2$

where, $K C l$ $KCl$ = Potassium Chloride,
$O_{2}$ $O_2$ = Oxygen Gas
$K C l O_{3}$ $KClO_3$ = Potassium Chlorate

So, 6 moles of Oxygen Gas $O_{2}$ $O_2$ has been given to you.
To calculate the number of moles for potassium chlorate $K C l O_{3}$ $KClO_3$ , use the mole ratio.

Looking closely at the equation,
If 3 moles of $O_{2}$ $O_2$ gives you 2 moles of $K C l O_{3}$ $KClO_3$ because $3 \div 3 \times 2$ $3 -: 3 xx 2$ ,
then 6 moles of $O_{2}$ $O_2$ must give you:
$6 \div 3 \times 2$ $6 -: 3 xx 2$ = 4 moles of $K C l O_{3}$ $KClO_3$

To find the number of moles for another compound in the same equation:
number of moles given to you from compound 1 $\div$ $-:$ the front number for compound 1 $\times$ $xx$ the front number for another compound.

Therefore, 4 moles of potassium chlorate must be used to produce 6 moles of oxygen gas.

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How many moles of potassium chlorate must be used to produce 6 moles of oxygen gas?

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