In the decomposition of kclo3 how many moles of kcl are formed in the reaction that produces 0.05 moles of o2
You'd need 33.3 moles of potassium chlorate, KClO3, to produce that much oxygen.
All you need in order to answer this question is the balanced chemical equation for the decomposition reaction of potassium chlorate 2KClO3(s)→2KCl(s)+3O2(g)
Notice that you have a 2:3 mole ratio between potassium chlorate and oxygen gas, which means that, regardless of how many moles of the former react, you'll always produce 3/2 times more moles of the latter.
Since you know how many moles of oxygen you need the reaction to produce, you can work backwards and determine how many moles of potassium chlorate you need by using the same mole ratio
50molesO2 ⋅ 2 moles KClO3/2moles O2=33.3 moles KClO3
oops Helper. It's 0.05 mols O2. Your answer is for millimoles. Second, the question is about KCl and not KClO3. The answer, in moles, is the same of course because the coefficient for KCl and KClO3 are the same.
543.9 mol
To determine the number of moles of KCl formed in the decomposition of KClO3, we need to know the balanced chemical equation for the reaction.
The balanced equation for the decomposition of KClO3 is:
2 KClO3 -> 2 KCl + 3 O2
According to the balanced equation, 2 moles of KCl are formed for every 2 moles of KClO3 decomposed. We can use this ratio to find out how many moles of KCl are formed when 0.05 moles of O2 are produced.
Using stoichiometry, we can set up a proportion to solve for the number of moles of KCl:
(0.05 moles O2) / (3 moles O2) = (x moles KCl) / (2 moles KCl)
Cross-multiplying gives us:
0.05 moles O2 * 2 moles KCl = 3 moles O2 * x moles KCl
0.1 moles KCl = 3x moles KCl
Dividing both sides by 3 gives us:
0.1 moles KCl / 3 = x moles KCl
So, x = 0.0333 moles KCl
Therefore, 0.05 moles of O2 would produce approximately 0.0333 moles of KCl in the decomposition of KClO3.
first, use proper capitalization to avoid confusion. KClO3
Next, use the balanced equation to show how moles of KCl and O2 are related.