Q22.81CP - Chemistry: Molecular Nature Of Matter And Change

Question

World production of chromite $\left({\text{FeCr}}_{\text{2}}{\text{O}}_{\text{4}}\right)$ , the main ore of chromium, was  ${\mathbf{\text{1.5×10}}}^{\mathbf{\text{7}}}$ metric tons in $\mathbf{2003}$ . To isolate chromium, a mixture of chromite and sodium carbonate is heated in air to form sodium chromate, iron(III) oxide, and carbon dioxide. The sodium chromate is dissolved in water, and this solution is acidified with sulfuric acid to produce the less soluble sodium dichromate. The sodium dichromate is filtered out and reduced with carbon to produce chromium(III) oxide, sodium carbonate, and carbon monoxide. The chromium(III) oxide is then reduced to chromium with aluminium metal. (a) Write balanced equations for each step. (b) What mass of chromium (in kg) could be prepared from the $\mathbf{2003}$  world production of chromite?

Step-by-Step Solution

Answer

a) Chemical equation is,

 $$\begin{gathered} {\text{4FeCr}}_{\text{2}}{\text{O}}_{\text{4}}{\text{(s) + 8Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{(s) + 7O}}_{\text{2}}\text{(g)}\to {\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + 8CO}}_{\text{2}}{\text{(g) + 8Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(s)} \\ {\text{2Na}}_{\text{2}}{\text{CrO}}_{\text{4}}{\text{(aq) + H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\to {\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{\text{(s) + Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{(aq) + H}}_{\text{2}}\text{O(l)} \\ {\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{(s) + 2C(s)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(aq) + CO(g)} \\ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 2Al(s)}\to {\text{2Cr(s) + Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)} \end{gathered}$$

b) The mass is$6.97\times {10}^{9}kg$  .

Step 1: Define mass

The amount of matter in a thing is measured by its mass. The most common units of mass are grammes (g) and kilogrammes (kg). The quantity of matter measured by mass is independent of its location in the cosmos and the gravitational force acting on it.

Step 2: Explanation

a) The following chemical equations can be used to describe the process:

 $$\begin{gathered} {\text{4FeCr}}_{\text{2}}{\text{O}}_{\text{4}}{\text{(s) + 8Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{(s) + 7O}}_{\text{2}}\text{(g)}\to {\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + 8CO}}_{\text{2}}{\text{(g) + 8Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(s)} \\ {\text{2Na}}_{\text{2}}{\text{CrO}}_{\text{4}}{\text{(aq) + H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\to {\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{\text{(s) + Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{(aq) + H}}_{\text{2}}\text{O(l)} \\ {\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{(s) + 2C(s)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(aq) + CO(g)} \\ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 2Al(s)}\to {\text{2Cr(s) + Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)} \end{gathered}$$

Step 3: Explanation

b) In$2003$  , the global production of chromite, {$\left.{\text{FeCr}}_{\text{2}}{\text{O}}_{\text{4}}\right\}$ , was  ${\text{1.5×10}}^{\text{7}}$ metric tonnes. Let's see how much chromium that amount of chromite could produce:

$$\begin{gathered} \text{m}\left({\text{FeCr}}_{\text{2}}{\text{O}}_{\text{4}}\right){\text{ = 1.5×10}}^{\text{7}}\text{t} \\ {\text{ = 1.510}}^{\text{13}}\text{kg} \\ \text{M}\left({\text{FeCr}}_{\text{2}}{\text{O}}_{\text{4}}\right){\text{ = 223.8348gmol}}^{\text{ - 1}} \\ \text{n}\left({\text{FeCr}}_{\text{2}}{\text{O}}_{\text{4}}\right)\text{ = }\frac{\text{m}\left({\text{FeCrO}}_{\text{4}}\right)}{\text{M}\left({\text{FeCrO}}_{\text{4}}\right)} \\ \text{n}\left({\text{FeCr}}_{\text{2}}{\text{O}}_{\text{4}}\right)\text{ =}\frac{{\displaystyle {\text{1.5×10}}^{\text{16}}\text{g}}}{{\displaystyle \text{223.8348}{\mathrm{gmol}}^{\text{ - 1}}}} \\ \text{n}\left({\text{FeCr}}_{\text{2}}{\text{O}}_{\text{4}}\right){\text{ = 6.7×10}}^{\text{10}}\text{mol} \end{gathered}$$

Since each chromite molecule contains  $\text{2 Cr}$ atoms, the total amount of $Cr$  is:

 $$\begin{gathered} \text{n(Cr) = 2n×}\left({\text{FeCr}}_{\text{2}}{\text{O}}_{\text{4}}\right) \\ {\text{n(Cr) = 1.3410}}^{\text{11}}\text{ mol} \end{gathered}$$

We can simply compute  $Cr$'s mass now:

 $$\begin{gathered} \text{m(Cr) = n(Cr)×M(Cr)} \\ {\text{m(Cr) = 1.34×10}}^{\text{11}}{\text{mol×51.9961gmol}}^{\text{ - 1}} \\ {\text{m(Cr) = 6.97×10}}^{\text{12}}\text{g} \\ {\text{ = 6.97×10}}^{\text{9}}\text{ kg} \end{gathered}$$

Therefore, the mass is ${\text{6.97×10}}^{\text{9}}\text{ kg}$ .