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Which of the following statements is/are correct when a mixture of NaCl and K₂Cr₂O₇ is gently warmed with conc. H₂SO₄?

Options:
a. A deep red vapor is evolved.
b. The red vapor when passed into NaOH solution gives a yellow solution of Na₂CrO₄.
c. Oxygen gas is evolved.
d. Chromyl chloride is formed.

User Kadepeay
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1 Answer

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Final answer:

When a mixture of NaCl and K₂Cr₂O₇ is gently warmed with conc. H₂SO₄, chromyl chloride is formed and the red vapor produced reacts with NaOH to give a yellow solution of Na₂CrO₄.

Step-by-step explanation:

When a mixture of NaCl and K₂Cr₂O₇ is gently warmed with conc. H₂SO₄, several reactions occur.

One of the correct statements is that the red vapor produced when the mixture is passed into NaOH solution gives a yellow solution of Na₂CrO₄. This is because K₂Cr₂O₇ reacts with conc. H₂SO₄ to form chromyl chloride, CrO₂Cl₂, which is a red vapor. When this vapor reacts with NaOH, it forms Na₂CrO₄, which is a yellow solution. The reaction can be represented as:

K₂Cr₂O₇ + 4NaOH + 6H₂SO₄ → 2CrO₂Cl₂ + 2KHSO₄ + 4H₂O + 4Na₂SO₄

Another correct statement is that chromyl chloride, CrO₂Cl₂, is formed during this process. Chromyl chloride is an orange-red compound that is formed when K₂Cr₂O₇ reacts with conc. H₂SO₄. The formation of chromyl chloride is responsible for the red color of the vapor produced.

In summary, when a mixture of NaCl and K₂Cr₂O₇ is gently warmed with conc. H₂SO₄, the red vapor produced is chromyl chloride, which reacts with NaOH to form a yellow solution of Na₂CrO₄.

User BLeB
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