A 3.8-mol sample of KClO3 was decomposed according to the equation. How many moles of O2 are formed assuming 100% yield?

Chemistry

1 answer:

kari74 [83]2 years ago

3 0

Answer:

5.7 moles of O2

Explanation:

We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:

2KClO3 —> 2KCl + 3O2

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.

Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.

Thus, 5.7 moles of O2 were obtained from the reaction.

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Explanation:

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<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$molarity=\frac{number of moles}{volume}$