91Ó°ÊÓ

In the given reaction, 5 mol of ${\mathit{O}}_{\mathbf{2}}$are formed from 4 mol of $\mathit{K}\mathit{N}{\mathit{O}}_{\mathbf{3}}$

Thus,

role="math" localid="1656741424651" $$\begin{gathered} \mathbf{5}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{O}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{K}\mathit{N}{\mathit{O}}_{\mathbf{3}} \\ \mathbf{1}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{O}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{4}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{KNO}}_{\mathbf{3}}}{\mathbf{5}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{K}\mathit{N}{\mathit{O}}_{\mathbf{3}} \end{gathered}$$

Therefore,1mol of ${\mathit{O}}_{\mathbf{2}}$is formed by 0.8 mol of $\mathit{K}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{.}$

The number of moles is calculated by the mass and Molar mass. The relationship between the number of moles, mass, and molar mass is given below.

$\mathit{N}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}$

The mass of ${\mathit{O}}_{\mathbf{2}}$= 56.6kg = 56600 g.

The molar mass of ${\mathit{O}}_{\mathbf{2}}$=31.999 g/mol

Thus, the number of moles of ${\mathit{O}}_{\mathbf{2}}$is:

$\overline{)\begin{array}{ccc}\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{O}}_{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{mass}}{\mathbf{Molar}\mathbf{}\mathbf{mass}}\\ & \mathbf{=}& \frac{\mathbf{56600}\mathbf{}\mathbf{g}}{\mathbf{31}\mathbf{.}\mathbf{999}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}}\\ & \mathbf{=}& \mathbf{1768}\mathbf{.}\mathbf{80528}\mathbf{}\mathbf{mol}\\ & \mathbf{;}& \mathbf{1768}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{1}\mathbf{mol}\end{array}}$

The number of $KN{O}_3$is:

$\begin{array}{ccc}\mathbf{1}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}{\mathbf{O}}_{\mathbf{2}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{K}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}\\ \mathbf{1768}\mathbf{.}\mathbf{81}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}{\mathbf{O}}_{\mathbf{2}}& \mathbf{=}& \mathbf{1768}\mathbf{.}\mathbf{81}\mathbf{×}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{K}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}\\ & \mathbf{=}& \mathbf{1415}\mathbf{.}\mathbf{048}\mathbf{.}\mathbf{048}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{K}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}\end{array}$

Therefore, the number of moles of $KN{O}_3$is approximately $\mathbf{1}\mathbf{.}\mathbf{41}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{.}$

The mass of $KN{O}_3$is:

$\begin{array}{ccc}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{K}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}& \mathbf{=}& \mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{K}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}\mathbf{×}\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{K}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}\\ & \mathbf{=}& \mathbf{1415}\mathbf{.}\mathbf{048}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{×}\mathbf{101}\mathbf{.}\mathbf{1032}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{m}\mathbf{o}\mathbf{l}\\ & \mathbf{=}& \mathbf{143065}\mathbf{.}\mathbf{881}\mathbf{g}\end{array}$

Therefore, the mass of $KN{O}_3$is approximately $\mathbf{1}\mathbf{.}\mathbf{43}\mathbf{×}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{g}$