Question

A mixture containing ${\rm{KCl}}{{\rm{O}}_3}$,${\rm{KHC}}{{\rm{O}}_3}$,${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$and ${\rm{KCl}}$ was heated , producing ${\rm{C}}{{\rm{O}}_2}$, ${{\rm{O}}_2}$and ${{\rm{H}}_2}{\rm{O}}$ gases according to the following equations:
${\rm{2}}\,{\rm{KCl}}{{\rm{O}}_{\rm{3}}}\,{\rm{(s)}} \to \,{\rm{2}}\,{\rm{KCl(s) + }}\,{\rm{3}}{{\rm{O}}_{\rm{2}}}\,\,$
${\rm{2}}\,{\rm{KHC}}{{\rm{O}}_{\rm{3}}}\,{\rm{(s)}} \to \,{\rm{2}}\,{{\rm{K}}_{\rm{2}}}{\rm{O(s) + }}\,\,{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\,{\rm{ + }}\,2\,{\rm{C}}{{\rm{O}}_{\rm{2}}}\,{\rm{(g)}}\,$
$\,{{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}{\rm{(s)}} \to \,{{\rm{K}}_{\rm{2}}}{\rm{O(s) + }}\,{\rm{C}}{{\rm{O}}_{\rm{2}}}\,{\rm{(g)}}\,$
The ${\rm{KCl}}$ does not react under the conditions of the reaction. If ${\rm{100}}{\rm{.0}}\,{\rm{g}}$of the mixture produces ${\rm{1}}{\rm{.80}}\,{\rm{g}}$of ${{\rm{H}}_2}{\rm{O}}$, ${\rm{13}}{\rm{.20}}\,{\rm{g}}$of ${\rm{C}}{{\rm{O}}_2}$and ${\rm{4}}{\rm{.0 g}}$of ${{\rm{O}}_2}$, what was the composition of the original mixture?

Answer 1

As we know that the reactions are balanced reactions which obey the law of conservation of mass. This question can be calculated if we could know that one mole of reactant produces how much gram of product.

Complete step-by-step solution: Now, coming on our given question,
When two moles${\rm{(245}}\,{\rm{g)}}$of ${\rm{KCl}}{{\rm{O}}_3}$is heated, then three moles of dioxide ${\rm{(96}}\,{\rm{g)}}$ is formed as;

{\rm{2}}\,{\rm{KCl}}{{\rm{O}}_{\rm{3}}}\,{\rm{(s)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,{\rm{2}}\,{\rm{KCl(s)}}\,\,\,{\rm{ + }}\,\,{\rm{3}}{{\rm{O}}_{\rm{2}}}\,\,\\\ {\rm{2}}\, \times (39 + 35.5 + 48)\,\,{\rm{g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\,3 \times 32\,){\rm{g}}\\\ {\rm{245}}\,{\rm{g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{96}}\,{\rm{g}} \end{array}$$ So, we are given as that $${\rm{4}}{\rm{.0 g}}$$of $${{\rm{O}}_2}$$is formed Then, $${\rm{96}}{\rm{.0 g}}$$of $${{\rm{O}}_2}$$is produced by $${\rm{245 g}}$$of $${\rm{KCl}}{{\rm{O}}_3}$$then $${\rm{4}}{\rm{.0 g}}$$of $${{\rm{O}}_2}$$is formed by $${\rm{ = }}\,\dfrac{{245\,{\rm{gm}}}}{{96\,{\rm{gm}}}}{\rm{ \times }}\,4\,{\rm{gm}}$$ $${\rm{KCl}}{{\rm{O}}_3}$$ Therefore, $$1\,0.2\,{\rm{g}}$$of $${\rm{KCl}}{{\rm{O}}_3}$$produces $${\rm{4}}{\rm{.0 g}}$$of $${{\rm{O}}_2}$$. Now, come on the second given equation, we will follow the similar steps as When two mole$${\rm{(200}}\,{\rm{g)}}$$of $${\rm{KHC}}{{\rm{O}}_3}$$is heated , then one mole of water$${\rm{(18}}\,{\rm{g)}}$$ is formed as $$\begin{array}{l} {\rm{2}}\,{\rm{KHC}}{{\rm{O}}_{\rm{3}}}\,{\rm{(s)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,{\rm{2}}\,{{\rm{K}}_{\rm{2}}}{\rm{O(s) + }}\,\,{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\,{\rm{ + }}\,2\,{\rm{C}}{{\rm{O}}_{\rm{2}}}\,{\rm{(g)}}\,\\\ 2\, \times \,(\,39 + 1 + 12 + 48)\,{\rm{g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2 + 16)\,{\rm{g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\ {\rm{200}}\,{\rm{g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,18\,{\rm{g}} \end{array}$$ So, we are given as that $${\rm{1}}{\rm{.80}}\,{\rm{g}}$$of $${{\rm{H}}_2}{\rm{O}}$$is formed Then, $${\rm{18}}{\rm{.0 g}}$$of $${{\rm{H}}_2}{\rm{O}}$$ is produced by $${\rm{200 g}}$$of $${\rm{KHC}}{{\rm{O}}_3}$$then $${\rm{1}}{\rm{.80}}\,{\rm{g}}$$of $${{\rm{H}}_2}{\rm{O}}$$is formed by $${\rm{ = }}\,\dfrac{{200\,{\rm{gm}}}}{{{\rm{18}}\,{\rm{gm}}}}{\rm{ \times }}\,1.80\,{\rm{gm}}$$ $${\rm{KHC}}{{\rm{O}}_3}$$ Therefore, $${\rm{2}}\,0\,{\rm{g}}$$of $${\rm{KHC}}{{\rm{O}}_3}$$produces $${\rm{1}}{\rm{.80}}\,{\rm{g}}$$of $${{\rm{H}}_2}{\rm{O}}$$. Now, come on next reaction. When one mole$${\rm{(138}}\,{\rm{g)}}$$of $${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$$ is heated , then one mole of carbon dioxide$${\rm{(44}}\,{\rm{g)}}$$ is formed as $$\begin{array}{l} \,{{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}{\rm{(s)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,{\rm{2}}\,{{\rm{K}}_{\rm{2}}}{\rm{O(s) + }}\,{\rm{C}}{{\rm{O}}_{\rm{2}}}\,{\rm{(g)}}\,\\\ (\,2\, \times \,39\, + 12 + 48)\,{\rm{g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(12 + 32)\,{\rm{g}}\\\ {\rm{138}}\,{\rm{g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,44\,{\rm{g}} \end{array}$$ So, we are given as that $${\rm{13}}{\rm{.20}}\,{\rm{g}}$$of $${\rm{C}}{{\rm{O}}_2}$$is formed Then, $${\rm{44}}{\rm{.0 g}}$$of $${\rm{C}}{{\rm{O}}_2}$$ is produced by $${\rm{138}}\,{\rm{g}}$$of $${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$$then $${\rm{13}}{\rm{.20}}\,{\rm{g}}$$of $${\rm{C}}{{\rm{O}}_2}$$is formed by $${\rm{ = }}\,\dfrac{{138\,{\rm{gm}}}}{{44\,{\rm{gm}}}}{\rm{ \times }}\,13.20\,{\rm{gm}}$$$${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$$ Therefore, $${\rm{41}}{\rm{.4g}}$$of$${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$$ produces $${\rm{13}}{\rm{.20}}\,{\rm{g}}$$$${\rm{C}}{{\rm{O}}_2}$$. Now, the composition of $${\rm{KCl}}$$is calculated as – $ {\rm{composition}}\,{\rm{of}}\,{\rm{KCl}}\,{\rm{ = }}\,{\rm{total}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{mixture - }}\,{\rm{sum}}\,{\rm{of}}\,{\rm{calculated}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{reactants}}\\\ {\rm{composition}}\,{\rm{of}}\,{\rm{KCl}}\,{\rm{ = }}\,{\rm{100}}\,{\rm{g - }}\,{\rm{(10}}{\rm{.2}}\,{\rm{g + }}\,{\rm{20}}\,{\rm{g + }}\,{\rm{41}}{\rm{.4}}\,{\rm{g)}}\\\ {\rm{composition}}\,{\rm{of}}\,{\rm{KCl}}\,{\rm{ = }}\,2{\rm{8}}{\rm{.4}}\,{\rm{g}} $ **Therefore, the composition of mixture is: $${\rm{KCl}}{{\rm{O}}_{\rm{3}}}\,{\rm{ = }}\,1{\rm{0}}{\rm{.2}}\,{\rm{g}}$$ $${\rm{KHC}}{{\rm{O}}_3}\, = \,{\rm{2}}\,0\,{\rm{g}}$$ $${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\, = \,{\rm{41}}{\rm{.4g}}$$ $${\rm{KCl}}\,{\rm{ = }}\,2{\rm{8}}{\rm{.4}}\,{\rm{g}}$$** **Note:** The above method is the easy method to calculate the mass of reactant or product. Only we have to know the balanced reaction. Some questions do not contain any reaction so we have to use the law of conservation of mass to balance the reacting species and products.