Question

How many grams of oxygen can be produced if you have \[5.0\] grams of \[KCl{{O}_{3}}\] in the following reaction?
\[2KCl{{O}_{3}}\text{ }\to \text{ }2KCl\text{ }+\text{ }3{{O}_{2}}\]

Answer 1

Hint:We know that the Stoichiometry always allow to create prediction about outcome of chemical reaction and creating a grateful prediction which is one of main goal of science with other being a ability to need to explain phenomena so that we can observe in natural phenomenon.

Complete solution:
Here initially we start with $1$ mol of potassium chlorate which give almost 3 mole of oxygen from the given chemical equation, thus it can be written as a \[\dfrac{3\text{ }mol\text{ }{{O}_{2}}}{1\text{ }mol\text{ }KCl{{O}_{3}}}\text{ }\] which is given in stoichiometric ratio.
So in order to convert gram of compound \[KCl{{O}_{3}}\] to moles we have to follow these steps
We have given that $5g$ of \[KCl{{O}_{3}}\] and this is as same as
\[5\text{ }g\text{ }KCl{{O}_{3}}\text{ }\times \text{ }\left( \dfrac{1\text{ }mol\text{ }KCl{{O}_{3}}}{122.55\text{ }g} \right)\]
Where g cancel out in numerator as well as in denominator to get a yield of \[\left( \dfrac{5}{122.55} \right)\text{ }=\text{ }0.040799674\text{ }mol\]
So in order to obtain mole of ${{O}_{2}}$ released in chemical reaction we have stoichiometric ratio which we know that is used here
Using stoichiometry to calculate mole of ${{O}_{2}}$ released is given by:
\[\left( \dfrac{3\text{ }mol\text{ }{{O}_{2}}}{1\text{ }mol\text{ }KCl{{O}_{3}}} \right)\times 0.04079\text{ }mol\text{ }KCl{{O}_{3}}\] where a mol \[KCl{{O}_{3}}\] cancel out in numerator as well as denominator with a yield of \[0.122399021\text{ }mol\text{ }of\text{ }{{O}_{2}}\]
After having done with stoichiometry we have to use molar mass of ${{O}_{2}}$ so that we have to convert mol into $g$ ;
\[\dfrac{32\text{ }g\text{ }{{O}_{2}}}{1mol\text{ }{{O}_{2}}}\text{ }\left( molar\text{ }mass \right)\] from periodic element. Each of the provided O weights \[\text{16 }\dfrac{g}{mol}\] also O2 has two atom and thus we have molar mass and it would be \[32\text{ }\dfrac{g}{mol}\]
Therefore, \[0.12239\text{ }mol\text{ }{{O}_{2}}\times \left( \dfrac{32\text{ }g}{mol\text{ }{{O}_{2}}} \right)\] (mol ${{O}_{2}}$ cancels out in numerator as well as in denominator we have a yield of \[3.91679\]
Thus the mass of ${{O}_{2}}$ released would be $\approx 4.0g$.

Note:Note that we have the prediction of mass of the product of chemical reaction given and if the given has a starting mass of reactant. Also to determine the optimal ratio of a reactant for chemical reaction so that all of the given reactants are used at their fully efficiency.