Question

In the decomposition of potassium chlorate, how many moles of potassium chlorate are needed to produce 50 moles of oxygen gas?

Answer 1

To determine the no. of moles and also weight (in grams) of any reactant or product we’ll first have to write the balanced chemical equation of the reaction. After writing the reaction, the no. of moles of each required will be known clearly.

Complete answer:
The reaction given to us is the decomposition of Potassium Chlorate. The decomposition of Potassium chlorate gives us Potassium Chloride (KCl) and Oxygen. The reaction can be given as:
$KCl{O_{3(s)}} \to KCl + {O_2}$
Balancing the above equation to balance the no. of oxygen, we obtain the final reaction as:
$2KCl{O_{3(s)}} \to 2KC{l_{(s)}} + 3{O_{2(g)}}$
From the above reaction we can tell that 2 moles of potassium chlorate will produce 2 moles of KCl and 3 moles of Dioxygen.
Comparing the mole ratio to find out the no. of moles required to produce 50 moles of Oxygen.
$2Moles$ of $KCl{O_3}$ $= 3Moles$ of ${O_2}$
$1mol$ of ${O_2}$ $= \dfrac{2}{3}mol$ of $KCl{O_3}$
We have to find the no. of moles of $KCl{O_3}$ required for 50 moles of Oxygen
$50moles$ of ${O_2}$ $= \dfrac{2}{3} \times 50 = 33.33mol$ of $KCl{O_3}$
Therefore, the required answer is 33.33 moles of Potassium Chlorate.

Additional Information:
In the laboratory, heating of potassium chlorate and manganese dioxide evolves oxygen at a very high temperature of 650K. At this temperature, Potassium Chlorate melts into Potassium Perchlorate, which further decomposes to give Oxygen at a higher temperature.

Note:
The stoichiometric relations give us the quantitative relationships between the reactants and products. To know the stoichiometric relations a balanced chemical equation is required. If we are asked to find the mass $KCl{O_3}$ we would have used the formula $mass(g) = moles \times M.{M_{KCl{O_3}}}$. For this the molar mass is required to be found out.