Question

$ 367.5g $ of $ KCl{O_3} $ is heated. How many grams of $ KCl $ and $ {O_2} $ is produced?
 $ 2KCl{O_3}\xrightarrow{{}}2KCl + 3{O_2} $
Solve using mass-mass relation by stoichiometry.

Answer 1

Hint: We know that stoichiometry is a branch of chemistry that deals with the relationship between relative quantities of substances that are involved in a chemical reaction. To find the mass-to-mass relation between the reactants of a product, you should at first establish a relationship between the moles and balance the equation based on it. We can calculate the moles of reactant and use stoichiometry to find the moles and thus, the weight of the products.

Complete Step by Step answer:
We have the chemical equation,
 $ 2KCl{O_3}\xrightarrow{{}}2KCl + 3{O_2} $
Which seems to be perfectly balanced with 2 potassium atoms, 2 chlorine atoms and 6 oxygen atoms on the reactant and product side of the reaction.
Now that we have a completely balanced equation, let us find the moles balanced in this reaction. Let us start by calculating the molar mass of the reactant.
Molar mass of $ KCl{O_3} $ = Atomic mass of $ K $ + Atomic mass of $ Cl $ + 3(Atomic mass of $ O $ )
= $ 39 + 35.5 + 3 \times 16 $
 $ = {\text{ }}122.5{\text{g}} $
Now we know that I mole of $ KCl{O_3} $ = 122.5 g
Thus, 1g of $ KCl{O_3} $ = $ \frac{{1{\text{mole}}}}{{122.5g}} $
We have been given $ 367.5{\text{ g}} $ of $ KCl{O_3} $ , which will be,
 $ 367.5{\text{ g}} $ of $ KCl{O_3} $ = $ \frac{{1{\text{mole}}}}{{122.5g}} \times 367.5 = 3{\text{mole}} $
Thus, we have 3 moles of $ KCl{O_3} $ which on heating becomes $ KCl $ and $ {O_2} $ . However, from the balanced chemical equation we know that 2 moles of $ KCl{O_3} $ gives 2 moles of $ KCl $ and 3 moles of $ {O_2} $ .
For $ KCl $
Since 2 moles of $ KCl{O_3} $ produces 2 moles of $ KCl $ , 3 moles of $ KCl{O_3} $ would produce $ 3 \times \frac{2}{2} $ moles of $ KCl $ ,which is $ 3{\text{moles}} $ . We know that 1 mole of $ KCl $ = $ 39 + 35.5 = 74.5g $ , similarly for 3 moles of $ KCl $ , the molar mass produced would be $ \left( {3 \times 74.5} \right)g = 223.5g $ .
For $ {O_2} $
From the balanced equation we know that 2 moles of $ KCl{O_3} $ produces 3 moles of $ {O_2} $ , thus 3 moles of $ KCl{O_3} $ we would get, $ \left( {\frac{{3 \times 3}}{2}} \right){\text{moles}} $ , which would be $ 4.5{\text{moles}} $ .
We also know that the molar mass for 1 mole of oxygen is $ \left( {2 \times 16} \right){\text{g}} = 32{\text{g}} $ , thus for $ 4.5{\text{moles}} $ of oxygen, the molar mass produced would be, $ \left( {32 \times 4.5} \right)g = 144g $ .
Thus, the mass of $ KCl $ and $ {O_2} $ produced when 3 moles of $ KCl{O_3} $ heated would be $ 235.5{\text{g}} $ and $ 144{\text{g}} $ respectively.

Note:
For stoichiometric equations it is always required that the equations are perfectly balanced before the mass-to-mass ratio or moles ratio is calculated. To balance equations perfectly alter the coefficients of compounds without changing the molecular formula.