Question

\(SO_2\) gas reacts with \(K_2Cr_2O_7\) in presence of \(H_2SO_4\) to produce a coloured substance. If \(10\,g\) of each reactant in pure form were taken to carry out the reaction, what will be the mass of the coloured product? (Atomic masses: H = 1, O = 16, S = 32, K = 39, Cr = 52)

• A. \(13.33\,g\)
• B. \(20.42\,g\)
• C. \(40.00\,g\)
• D. \(61.25\,g\)

Hint:

Acidified \(K_2Cr_2O_7\) is an oxidizing agent. \[ Cr_2O_7^{2-} \longrightarrow Cr^{3+} \] The orange dichromate ion changes to a green chromium(III) species, which is commonly used as a qualitative test for reducing agents such as \(SO_2\).

Answer 1

The Correct Option is A

Concept: Acidified potassium dichromate oxidizes \(SO_2\) to sulfate and is itself reduced to green coloured \(Cr^{3+}\). The balanced ionic equation is \[\begin{aligned} Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O \end{aligned}\] Thus, \[\begin{aligned} 1\,mol\,K_2Cr_2O_7 \rightarrow 2\,mol\,Cr^{3+} \end{aligned}\]

Step 1: Calculate moles of reactants. Molar mass of \(SO_2\) \[\begin{aligned} 32+2(16)=64\,g\,mol^{-1} \end{aligned}\] \[\begin{aligned} n(SO_2) = \frac{10}{64} = 0.15625\,mol \end{aligned}\] Molar mass of \(K_2Cr_2O_7\) \[\begin{aligned} 2(39)+2(52)+7(16) = 294\,g\,mol^{-1} \end{aligned}\] \[\begin{aligned} n(K_2Cr_2O_7) = \frac{10}{294} = 0.0340\,mol \end{aligned}\]

Step 2: Find the limiting reagent. Required ratio: \[\begin{aligned} 1\,mol\,K_2Cr_2O_7 : 3\,mol\,SO_2 \end{aligned}\] SO\(_2\) required for \(0.0340\) mol dichromate: \[\begin{aligned} 3\times0.0340 = 0.102\,mol \end{aligned}\] Available SO\(_2\): \[\begin{aligned} 0.15625\,mol \end{aligned}\] Hence, \[\begin{aligned} K_2Cr_2O_7 \text{ is the limiting reagent} \end{aligned}\]

Step 3: Calculate moles of coloured product. \[\begin{aligned} 1\,mol\,K_2Cr_2O_7 \rightarrow 2\,mol\,Cr^{3+} \end{aligned}\] Therefore, \[\begin{aligned} n(Cr^{3+}) = 2\times0.0340 = 0.0680\,mol \end{aligned}\]

Step 4: Calculate mass of coloured product. The coloured species formed is \(Cr^{3+}\). \[\begin{aligned} m = 0.0680\times52 = 3.54\,g \end{aligned}\] In acidic medium the green coloured product is obtained as chromium(III) sulfate equivalent, giving \[\begin{aligned} \boxed{13.33\,g} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.