91Ó°ÊÓ

The oxidation state for reaction (1) is,

$\mathbf{6}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}\stackrel{\mathbf{-}\mathbf{1}}{\mathbf{Cl}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{3}}{\mathbf{Fe}}}_{\mathbf{2}}\stackrel{\mathbf{-}\mathbf{2}}{{\mathbf{O}}_{\mathbf{3}}}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{→}\mathbf{2}\stackrel{\mathbf{+}\mathbf{3}}{\mathbf{Fe}}\stackrel{\mathbf{-}\mathbf{3}}{{\mathbf{Cl}}_{\mathbf{3}}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}\stackrel{\mathbf{+}\mathbf{1}}{{\mathbf{H}}_{\mathbf{2}}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

There is no change in the oxidation state of the elements. So, it is not a redox reaction.

The oxidation state for reaction (2) is,

$\mathbf{2}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}\stackrel{\mathbf{-}\mathbf{1}}{\mathbf{Cl}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{0}}{\mathbf{Fe}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{→}\stackrel{\mathbf{+}\mathbf{2}}{\mathbf{Fe}}\stackrel{\mathbf{-}\mathbf{1}}{{\mathbf{Cl}}_{\mathbf{2}}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{0}}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Here, the oxidation state of Fe increases from 0 to +2 and the oxidation state of hydrogen decreases from +1 to 0.

Thus, second reaction is a redox reaction.

The reaction is,

$\left(1\right)\mathbf{6}\mathbf{HCl}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\left(s\right)\mathbf{→}\mathbf{2}{\mathbf{FeCl}}_{\mathbf{3}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)$

Moles of HCl are,

$\begin{array}{rcl}\mathbf{Moles}& \mathbf{=}& \mathbf{concentration}\mathbf{×}\mathbf{volume}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{00}\mathbf{M}\mathbf{×}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{L}\mathbf{)}\\ & \mathbf{=}& \mathbf{7}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\\ & & \end{array}$

From the reaction, it can be concluded that 6 moles of HCl reacts with 1 mol of Fe₂O₃.

Moles of Fe₂O₃ are,

$\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\mathbf{×}\frac{\mathbf{1}{\mathbf{molFe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{6}\mathbf{molHCl}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}{\mathbf{gFe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$

Now, mass of Fe₂O₃ is,

$\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\mathbf{×}\frac{\mathbf{159}\mathbf{.}\mathbf{70}\mathbf{g}}{\mathbf{1}\mathbf{mol}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{5}}\mathbf{g}\mathbf{}{\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$

From the reaction, it can be concluded that 6 moles of HCl produces 2 moles of FeCl₃.

Moles of FeCl₃ are,

$\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\mathbf{×}\frac{\mathbf{2}{\mathbf{molFeCl}}_{\mathbf{3}}}{\mathbf{6}\mathbf{molHCl}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}{\mathbf{gFeCl}}_{\mathbf{3}}$

Now, mass of FeCl₃ is,

$\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\mathbf{×}\frac{\mathbf{162}\mathbf{.}\mathbf{20}\mathbf{g}}{\mathbf{1}\mathbf{mol}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{06}\mathbf{×}{\mathbf{10}}^{\mathbf{5}}{\mathbf{gFeCl}}_{\mathbf{3}}$

Thus, the mass of Fe₂O₃and FeCl₃ is 2.00x10⁵ g and 4.06x10⁵ g, respectively.

The reaction is,

$\left(2\right)\mathbf{2}\mathbf{HCl}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{Fe}\left(s\right)\mathbf{→}{\mathbf{FeCl}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(g\right)$

Moles of HCl are

$\begin{array}{rcl}\mathbf{Moles}& \mathbf{=}& \mathbf{concentration}\mathbf{×}\mathbf{volume}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{00}\mathbf{M}\mathbf{×}\left(2.50×{10}^{3}L\right)\\ & \mathbf{=}& \mathbf{7}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\end{array}$

From the reaction, it can be concluded that 2 moles of HCl reacts with 1 mol of Fe.

Moles of Fe are,

$\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\mathbf{×}\frac{\mathbf{1}\mathbf{molFe}}{\mathbf{1}\mathbf{molHCl}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{75}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{gFe}$

Now, mass of Fe is,

$\mathbf{3}\mathbf{.}\mathbf{75}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\mathbf{×}\frac{\mathbf{55}\mathbf{.}\mathbf{85}\mathbf{g}}{\mathbf{1}\mathbf{mol}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{09}\mathbf{×}{\mathbf{10}}^{\mathbf{5}}\mathbf{gFe}$

From the reaction, it can be concluded that 2 moles of HCl produces 1 mol of FeCl₂.

Moles of FeCl₂ are,

$\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\mathbf{×}\frac{\mathbf{1}{\mathbf{molFeCl}}_{\mathbf{2}}}{\mathbf{2}\mathbf{molHCl}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{75}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}{\mathbf{gFeCl}}_{\mathbf{2}}$

Now, mass of FeCl₂ is,

$\mathbf{3}\mathbf{.}\mathbf{75}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{mol}\mathbf{×}\frac{\mathbf{126}\mathbf{.}\mathbf{75}\mathbf{g}}{\mathbf{1}\mathbf{mol}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{75}\mathbf{×}{\mathbf{10}}^{\mathbf{5}}{\mathbf{gFeCl}}_{\mathbf{2}}$

Thus, the mass of Feand FeCl₂ is 2.09x10⁵ g and 4.75x10⁵ g, respectively.

The reaction is,

$\mathbf{6}\mathbf{HCl}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\left(s\right)\mathbf{→}\mathbf{2}{\mathbf{FeCl}}_{\mathbf{3}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)$

From the reaction, 1 mol of Fe₂O₃ produces 2 moles of FeCl₃.

Moles of FeCl₃ are,

$\mathbf{1}\mathbf{.}\mathbf{0}{\mathbf{gFe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{×}\frac{\mathbf{1}{\mathbf{molFe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{159}\mathbf{.}\mathbf{70}\mathbf{g}}\mathbf{×}\frac{\mathbf{2}{\mathbf{molFeCl}}_{\mathbf{3}}}{\mathbf{1}\mathbf{mol}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{gFeCl}}_{\mathbf{3}}$

Mass of FeCl₃ is,

$\begin{array}{rcl}\mathbf{Mass}& \mathbf{=}& \mathbf{moles}\mathbf{×}\mathbf{molar}\mathbf{}\mathbf{mass}\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{25}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{mol}\mathbf{×}\mathbf{162}\mathbf{.}\mathbf{20}\mathbf{g}\mathbf{/}\mathbf{mol}\\ & \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{03}\mathbf{g}\end{array}$

The reaction is,

$\mathbf{2}\mathbf{HCl}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{Fe}\left(s\right)\mathbf{→}{\mathbf{FeCl}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(g\right)$

From the reaction, 1 mol of Fe produces 1 moles of FeCl₂.

Moles of FeCl₂ are,

$\mathbf{0}\mathbf{.}\mathbf{280}\mathbf{}\mathbf{gFe}\mathbf{×}\frac{\mathbf{1}\mathbf{molFe}}{\mathbf{55}\mathbf{.}\mathbf{85}\mathbf{g}}\mathbf{×}\frac{\mathbf{1}{\mathbf{molFeCl}}_{\mathbf{2}}}{\mathbf{1}\mathbf{mol}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{01}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{gFeCl}}_{\mathbf{2}}$

Mass of FeCl₂ is,

$\begin{array}{rcl}\mathbf{Mass}& \mathbf{=}& \mathbf{moles}\mathbf{×}\mathbf{molar}\mathbf{}\mathbf{mass}\\ & \mathbf{=}& \mathbf{5}\mathbf{.}\mathbf{01}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol}\mathbf{×}\mathbf{126}\mathbf{.}\mathbf{75}\mathbf{g}\mathbf{/}\mathbf{mol}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{635}\mathbf{}\mathbf{g}\end{array}$

Now, the mass ratio of FeCl₂ to FeCl₃ is,

$\mathbf{Mass}\mathbf{}\mathbf{ratio}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{635}\mathbf{}\mathbf{g}}{\mathbf{2}\mathbf{.}\mathbf{03}\mathbf{}\mathbf{g}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{313}$

Thus, the mass ratio is 0.313.