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

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.