How many moles of oxygen can be obtained by the decomposition of 7.5 mol of reactant in the following reaction?
2KClO3 -> 2KCl + 3O2
Here is the exact problem I've worked out. Start with moles (you can omit the equation step and converting grams to moles step).
http://www.jiskha.com/science/chemistry/stoichiometry.html
That would be 11.25 mol O2
Process:
7.5 mol KClO3 x 3 mol O2/2 mol KCOl3 = 11.25 mol O2
Note: Cancel the mol KCLO3
Well, let's put it this way: the reactant gave up its life, so the question is asking how many oxygen moles will arise from its decomposition.
Given that the reaction is 2KClO3 -> 2KCl + 3O2, we can see that for every 2 moles of KClO3 decomposed, 3 moles of O2 are produced.
So, if 2 moles of KClO3 produce 3 moles of O2, then we can set this up as a proportion:
2 moles KClO3 / 3 moles O2 = 7.5 moles KClO3 / x moles O2
We cross-multiply and solve for x, which gives us:
3 * 7.5 / 2 = x
So, it looks like x = 11.25 moles of O2! That means 7.5 moles of reactant will yield 11.25 moles of oxygen.
But hey, don't take my word for it! Double-check my math and calculations just to be sure, after all, I am just a Clown Bot!
To find the number of moles of oxygen obtained from the given reaction, you need to use the mole ratio from the balanced chemical equation.
According to the balanced equation:
2 KClO3 -> 2 KCl + 3 O2
From the equation, you can see that for every 2 moles of KClO3 that decompose, you will get 3 moles of O2. This means the mole ratio of KClO3 to O2 is 2:3.
Given that you have 7.5 moles of KClO3, you can set up a proportion to calculate the number of moles of O2:
(7.5 mol KClO3 / 2 mol KClO3) = (x mol O2 / 3 mol O2)
Simplifying the equation, you get:
x mol O2 = (7.5 mol KClO3 * 3 mol O2) / 2 mol KClO3
Evaluating the equation, you find:
x mol O2 = 11.25 mol O2
Therefore, the decomposition of 7.5 moles of KClO3 will yield 11.25 moles of O2.