Question

Limestone (CaCO₃) is used to remove acidic pollutants from smokestack flue gases. It is heated to form lime (CaO), which reacts with sulfur dioxide to form calcium sulfite. Assuming a 70.% yield in the overall reaction, what mass of limestone is required to remove all the sulfur dioxide formed by the combustion of 8.510⁴ kg of coal that is 0.33 mass % sulfur?

Short answer

You need to calculate the mass of limestone is required to remove all the sulfur dioxide formed by the combustion of 8.510⁴ kg of coal that is 0.33 mass % sulfur.

Step-by-step solution

Chemical reactions

Calcium carbonate decomposes and forms calcium oxide and carbon dioxide. The balanced equation is like

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\stackrel{\mathrm{\Delta }}{\to }\mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)......................................\left(1\right)$

Calcium oxide reacts with sulfur dioxide and forms calcium sulfite. The balanced equation is like

$\mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)\to {\mathrm{CaSO}}_3\left(\mathrm{s}\right)........................................\left(2\right)$

Sulfur reacts with oxygen and forms sulfur dioxide. The balanced equation is like

$\mathrm{S}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{SO}}_2\left(\mathrm{g}\right)........................................\left(3\right)$

Net balanced equation

Adding equation (1), (2) and (3); you will get

$\begin{array}{c}{\mathrm{CaCO}}_3\left(\mathrm{s}\right)\stackrel{\mathrm{\Delta }}{\to }\mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)\\ \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)\to {\mathrm{CaSO}}_3\left(\mathrm{s}\right)\\ \mathrm{S}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{SO}}_2\left(\mathrm{g}\right)\\ \overline{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)+\mathrm{S}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CaSO}}_3\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)}\end{array}$

Hence, the net balanced equation is

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)+\mathrm{S}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CaSO}}_3\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Calculation of moles of sulfur

According to the question

Mass of coal = 8.510⁴ kg = 8.510⁷ g

Mass percent of sulfur = 0.33%

Molar mass of sulfur = 32.06g/mol.

Now you can write,

Moles of sulfur

$\begin{array}{l}=\left[\frac{\mathrm{mass}\left(\mathrm{coal}\right)}{{\mathrm{molar}}_{\mathrm{mass}}\left(\mathrm{S}\right)}\right]\times \left[\frac{\mathrm{mass}\mathrm{\%}\left(\mathrm{S}\right)}{100\mathrm{\%}}\right]\\ =\left[\left(\frac{8.5\times {10}^7}{32.06}\right)\times \left(\frac{0.33\mathrm{\%}}{100\mathrm{\%}}\right)\right]\left(\mathrm{mol}\right)\\ =8749.22\left(\mathrm{mol}\right)\end{array}$

Hence, moles of sulfur are 8749.22mol.

Calculation of moles of CaCO3

According to the net balanced equation

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)+\mathrm{S}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CaSO}}_3\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

1 mol CaCO₃reacts with 1 mol sulfur.

Therefore, moles of CaCO₃

$\begin{array}{l}=\left(8749.22\times \frac{1}{1}\right)\left(\mathrm{mol}\right)\\ =8749.22\left(\mathrm{mol}\right)\end{array}$

Hence, moles of CaCO₃ are 8749.22mol.

Calculation of mass of CaCO3

According to the question,

Yield % = 70%

Molar mass of CaCO₃ = 100g/mol

Therefore, mass of CaCO₃

_{$\begin{array}{l}=\left[\left(8749.22\times \frac{100\mathrm{\%}}{70\mathrm{\%}}\right)\times 100\right]\left(\mathrm{g}\right)\\ =(1.24\times {10}^6)\left(\mathrm{g}\right)\\ =(1.24\times {10}^3)\left(\mathrm{kg}\right)\end{array}$}

Hence, mass of CaCO₃ = 1.2410³ kg.

Conclusion

The mass of limestone required to remove all the sulfur dioxide formed by the combustion of 8.510⁴ kg of coal that is 0.33 mass % sulfur is 1.2410³ kg.