Question

A sample of \[{\text{KCl}}{{\text{O}}_3}\] on decomposition yielded 448 mL of oxygen gas at NTP.
Calculate \[{\text{(i)}}\] weight of oxygen produced \[{\text{(ii)}}\] weight of \[{\text{KCl}}{{\text{O}}_3}\] originally taken \[{\text{(iii)}}\] weight of \[{\text{KCl}}\] produced.
\[{\text{2KCl}}{{\text{O}}_3}\xrightarrow{\Delta }{\text{ 2KCl + 3}}{{\text{O}}_2}\]
\[(K = 39,Cl = 35.5,{\text{ and }}O = 16)\]

Answer 1

Hint:For this question we must know the conversion of volume into number of moles on NTP that is normal condition of temperature and pressure. Molecular mass of all the elements are given to us, from that we can calculate their mass. Every product is produced according to stoichiometric or in simpler terms number of moles of reaction.
Formula used: \[{\text{number of moles at NTP}} = \dfrac{{V({\text{ml}})}}{{22400{\text{ml}}}}\]
\[{\text{number of moles }}n{\text{ = }}\dfrac{{{\text{mass of substance in gram}}}}{{{\text{molar mass}}}}\]

Complete step by step solution:
Whenever we want to find the product and reactant, then we will always proceed in stoichiometric ratio. As 2 moles of \[{\text{KCl}}{{\text{O}}_3}\] forms 2 moles of \[{\text{KCl}}\] and 3 moles of \[{{\text{O}}_2}\]so 1 mole of \[{\text{KCl}}{{\text{O}}_3}\] will form 1 mole of \[{\text{KCl}}\] and \[\dfrac{3}{2}\] moles of \[{{\text{O}}_2}\].
Volume of \[{{\text{O}}_2}\] is given so let us calculate the number of moles of \[{{\text{O}}_2}\]:
\[{\text{number of moles of }}{{\text{O}}_2}{\text{ }} = \dfrac{{{\text{448ml}}}}{{22400{\text{ml}}}} = 0.02{\text{ moles}}\]
Molar mass of \[{{\text{O}}_{\text{2}}}\] is \[{\text{16 + 16 = 32 gmo}}{{\text{l}}^{ - 1}}\]
Now we will calculate the mass of \[{{\text{O}}_{\text{2}}}\]
\[ \Rightarrow {\text{0}}{\text{.02 = }}\dfrac{{{\text{mass in (g)}}}}{{{\text{32 gmo}}{{\text{l}}^{ - 1}}}}\]
\[{\text{mass of }}{{\text{O}}_2} = 0.02 \times 32 = 0.64{\text{g}}\]
Molar mass of \[{\text{KCl}}{{\text{O}}_3}\] is \[{\text{39 + 35}}{\text{.5 + 3}} \times {\text{16 = 122}}{\text{.5 gmo}}{{\text{l}}^{ - 1}}\]
Since according to the reaction 2 moles of \[{\text{KCl}}{{\text{O}}_3}\] forms 3 mole of \[{{\text{O}}_{\text{2}}}\]. So mass of \[{\text{KCl}}{{\text{O}}_3}\] to produce 3 moles of \[{{\text{O}}_{\text{2}}}\] is:
 \[{\text{mass of KCl}}{{\text{O}}_3} = 2 \times 122.5 = 245{\text{g}}\]
But we have \[0.02\]moles of oxygen produce.
The required mass of \[{\text{KCl}}{{\text{O}}_3}\] for 3 moles of oxygen is 245 grams.
For \[0.02\] moles of oxygen mass of \[{\text{KCl}}{{\text{O}}_3}\] will be \[\dfrac{{245}}{3} \times 0.02 = 1.63{\text{ grams}}\]
According to the law of conservation of mass, mass is always conserved, that is mass of reactant and product will be the same. Hence,
\[{\text{mass of KCl}}{{\text{O}}_3} = {\text{ mass of KCl + mass of }}{{\text{O}}_2}\]
Rearranging we will get,
\[{\text{mass of KCl = mass of KCl}}{{\text{O}}_3}{\text{ - mass of }}{{\text{O}}_2}\]
\[ \Rightarrow {\text{mass of KCl }} = 1.63 - 0.64 = 0.993{\text{g}}\]

Note:
Stoichiometry is the minimum amount of reacting molecules that is required to run the reaction. For example in the given reaction 2 moles are required to produce the respective product. They are not the number of moles, because they do not tell us about the amount or reactant present. Instead stoichiometry tells about the minimum number of moles required.