What is the reaction between sulfur dioxide, #"SO"_2#, and potassium dichromate, #"K"_2 "Cr"_2"O"_7# ?

Answer 1

Well, we would assume that #"sulfur dioxide"# is oxidized to #"sulfur trioxide"#.

I think this is done industrially by #V_2O_5#.
But sulfur is oxidized to from #S(+IV)# to #S(+VI):#
#SO_2(g) +H_2O(l) rarr SO_3(g) +2H^(+) + 2e^(-)# #(i)#
And dichromate is reduced from #Cr(+VI)# to #Cr(+III):#
#Cr_2O_7^(2-) +14H^(+) +6e^(-)rarr 2Cr^(3+) +7H_2O(l)# #(ii)#
We add #3xx(i) + (ii)# to eliminate the electrons:
#Cr_2O_7^(2-) +8H^(+) +3SO_2(g) rarr 2Cr^(3+) +4H_2O(l)+3SO_3(g)#

Is the mass to charge ratio balanced? This is not my problem; it's yours.

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Answer 2

In presence of #"H"_2"SO"_4#,potassium dicromate reacts with #"SO"_2# as following.

#3"SO"_2# + #"K"_2"Cr"_2"O"_7# + #"H"_2"SO"_4# #-># #"K"_2"SO"_4# + #"Cr"_2("SO"_4)_3# + #"H"_2"O"#
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Answer 3

Chromium(III) sulfate (Cr2(SO4)3), sulfuric acid (H2SO4), and water (H2O) are the products of the reaction of sulfur dioxide (SO2) and potassium dichromate (K2Cr2O7) in an acidic medium.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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