One commercial system removes SO2 emissions from smoke at 95.0°C by the following set of balanced reactions: SO2(g) + Cl2 → SO2Cl2(g) SO2Cl2 + 2H2O → H2SO4 + 2HCl H2SO4 + Ca(OH)2 → CaSO4(s) + 2H2O Assuming the process is 95.0 % efficient, how many grams of CaSO4 may be produced from 100. g of SO2? (molar masses: SO2, 64.1 g/mol; CaSO4, 136 g/mol)
a. 87.2 g
b. 202 g
c. 44.8 g
d. 47.1 g
e. 212 g

[tex](1):SO_2(g)+Cl_2\rightarrow SO_2Cl_2(g)\\\\(2):SO_2Cl_2(g)+2H_2O(l)\rightarrow H_2SO_4(l)+2HCl(l)\\\\(3):H_2SO_4(l)+Ca(OH)_2(s)\rightarrow CaSO_4(s)+2H_2O(l)[/tex]

First we have to calculate the moles of [tex]SO_2[/tex].

[tex]\text{ Moles of }SO_2=\frac{\text{ Mass of }SO_2}{\text{ Molar mass of }SO_2}=\frac{100g}{64.1g/mole}=1.56moles[/tex]

Now we have to calculate the moles of [tex]CaSO_4[/tex].

From the given set of balanced reactions we conclude that,

As, the mole ratio of [tex]SO_2:SO_2Cl_2:H_2SO_4:CaSO_4[/tex] is, 1 : 1 : 1 : 1

So, the moles of [tex]CaSO_4[/tex] = moles of [tex]SO_2[/tex] = 1.56 moles

Now we have to calculate the mass of [tex]CaSO_4[/tex].

[tex]\text{ Mass of }CaSO_4=\text{ Moles of }CaSO_4\times \text{ Molar mass of }CaSO_4[/tex]

[tex]\text{ Mass of }CaSO_4=(1.56moles)\times (136g/mole)=212.16g[/tex]

As we are given that the process is 95.0 % efficient that means the amount we calculated is recovered.

Mass of [tex]CaSO_4[/tex] = [tex]212.16g\times 95\%=212.16g\times \frac{95}{100}=201.55g\approx 202g[/tex]

Therefore, the mass of [tex]CaSO_4[/tex] produced is 202 grams.