Question

Toxic Cr(VI) can be precipitated from an aqueous solution by bubbling SO2 through the
solution. How much SO 2 if required to treat 3.00 x 108 L of 5.50x10 2 mM Cr(VI)?
2Cro- +350 +4H
>
Cr2(SO4)3+2H20

Answer 1

Answer:

2.48 × 10⁴ mol

Explanation:

Step 1: Write the balanced equation

2 CrO₄²⁻ + 3 SO₂ + 4 H⁺ ⇒ Cr₂(SO₄)₃ + 2 H₂O

Step 2: Calculate the reacting moles of CrO₄²⁻

3.00 × 10⁸ L of 5.50 × 10⁻² mM CrO₄²⁻ react. The reacting moles are:

3.00 × 10⁸ L × 5.50 × 10⁻² × 10⁻³ mol/L = 1.65 × 10⁴ mol

Step 3: Calculate the reacting moles of SO₂

The molar ratio of CrO₄²⁻ to SO₂ is 2:3. The reacting moles of SO₂ are 3/2 × 1.65 × 10⁴ mol = 2.48 × 10⁴ mol.

Answer 2

1584 Kg of SO2 is required to treat 3.00 x 10^8 L of 5.50 x 10^-2 mM Cr(VI).

The equation of the reaction is;

2CrO4^2-(aq) + 3SO2(g) + 4H^+(aq) -------> Cr2(SO4)3(s) + 2H2O(l)

Number of moles of Cr(VI) = concentration × volume

concentration =  5.50 x 10^-2 mM or 5.5 x 10^-5 M

volume =  3.00 x 10^8 L

Number of moles of Cr(VI) = 5.5 x 10^-5 M × 3.00 x 10^8 L

= 1.65 x 10^4 moles

From the reaction equation;

2 moles of Cr(VI) reacts with 3 moles of SO2

1.65 x 10^4 moles reacts with  1.65 x 10^4 moles ×  3 moles/2 moles

= 24750 moles

Mass of SO2 = 24750 moles × 64 g/mol

Mass of SO2 = 1584 Kg of SO2

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