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

Step 1: Write the balanced reaction.
In acid, dichromate oxidises $SO_2$ and turns to green chromium(III) sulphate, the coloured product:
\[ K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O \]
So $1$ mole of dichromate reacts with $3$ moles of $SO_2$ to give $1$ mole of $Cr_2(SO_4)_3$.

Step 2: Find moles of each reactant.
Molar mass of $K_2Cr_2O_7 = 294$, so moles $= \dfrac{10}{294} = 0.034$.
Molar mass of $SO_2 = 64$, so moles $= \dfrac{10}{64} = 0.156$.

Step 3: Find the limiting reagent.
$0.034$ mol dichromate needs $3\times0.034 = 0.102$ mol $SO_2$. We have $0.156$ mol $SO_2$, which is more than enough, so dichromate is limiting.

Step 4: Find moles of product.
Since $1$ mole dichromate gives $1$ mole $Cr_2(SO_4)_3$, moles of product $= 0.034$.

Step 5: Find the molar mass of $Cr_2(SO_4)_3$.
\[ M = 2(52) + 3(32 + 4\times16) = 104 + 3(96) = 104 + 288 = 392 \]

Step 6: Find the mass of product.
\[ \text{mass} = 0.034 \times 392 \approx 13.33\ g \]
\[ \boxed{13.33\ g} \]