Question

Elemental chlorine was first produced by Carl Wilhelm Scheele in 1774 using the reaction of pyrolusite with sulfuric acid and sodium chloride:

4NaCl(s) + 2H₂SO₄(I) +MnO₂ (s)$\to$2NaSO₄ (s) + MnCl₂+ 2H₂O (I) + Cl₂(g)

Calculate the minimum mass of MnO₂ required to generate 5.32 L gaseous chlorine, measured at a pressure of 0.953 atm and a temperature of 33°C.

Short answer

The minimum mass of MnO₂ required for the reaction is 17.4 g.

Step-by-step solution

Step 1:

From the question,

Pressure (P) = 0.953 atm

Volume of chlorine gas (V) = 5.32 L

Temperature (T) = 33°C = 306 K

From ideal gas equation,

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \Rightarrow 0.953\mathrm{atm}\times 5.32\mathrm{L}=\mathrm{n}\times (0.082\mathrm{atm}.\mathrm{L}/\mathrm{Mol}.\mathrm{K})\times 306\mathrm{K} \\ \Rightarrow \mathrm{n}=\frac{0.953\times 5.32}{0.082\times 306}\mathrm{mol} \\ =0.2\mathrm{mol} \end{gathered}$$

Hence the number of moles of chlorine gas is 0.2 mol.

Step 2:

The molar mass of is M = 87g/mol

Let the number of moles of is “z” and mass of is “w”.

According to the stoichiometry, both the number of moles are same so n = z = 0.2 mol

$\mathrm{z}=\frac{\mathrm{w}}{\mathrm{M}}$

$$\begin{gathered} 0.2\mathrm{mol}=\frac{\mathrm{w}}{87\mathrm{g}/\mathrm{mol}} \\ \mathrm{w}=0.2\mathrm{mol}\times 87\mathrm{g}/\mathrm{mol} \\ =17.4\mathrm{g} \end{gathered}$$

Hence, the minimum mass of MnO₂required for the reaction is 17.4 g.