91Ó°ÊÓ

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 role="math" localid="1663315627514" $\mathbf{2003}$ world production of chromite?

Step by step solution

01

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.

02

Explanation

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

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

03

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:

$$\begin{gathered} \text{m(Cr) = n(Cr)×M(Cr)} \\ {\text{m(Cr) = 1.34×10}}^{\text{11}}{\text{³¾´Ç±ô×51.9961²µ³¾´Ç±ô}}^{\text{- 1}} \\ {\text{m(Cr) = 6.97×10}}^{\text{12}}\text{g} \\ {\text{= 6.97×10}}^{\text{9}}\text{kg} \end{gathered}$$

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

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!