Question

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

Answer 1

Answer : The correct option is, (b) 202 g

Explanation :

The given set of balanced reactions are:

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

First we have to calculate the moles of $SO_2$.

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

Now we have to calculate the moles of $CaSO_4$.

From the given set of balanced reactions we conclude that,

As, the mole ratio of $SO_2:SO_2Cl_2:H_2SO_4:CaSO_4$ is, 1 : 1 : 1 : 1

So, the moles of $CaSO_4$ = moles of $SO_2$ = 1.56 moles

Now we have to calculate the mass of $CaSO_4$.

$\text{ Mass of }CaSO_4=\text{ Moles of }CaSO_4\times \text{ Molar mass of }CaSO_4$

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

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

Mass of $CaSO_4$  = $212.16g\times 95\%=212.16g\times \frac{95}{100}=201.55g\approx 202g$

Therefore, the mass of $CaSO_4$  produced is 202 grams.