Fewpal
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K2Cr2O7 + SO2 + H2SO4 ⇒ Cr2(SO4)3 + K2SO4 + H2O

NOTE: Please explain the way you solve it in detail, that would be really helpful

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1 Answer

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In this case K2Cr2O7 is broken into three fragments namely K2O, Cr2O3 and nascent oxygen. This equation is immediately balanced. Then in step 2 and step 3, the oxides of K and Cr react with H2SO4 present in the medium to produce the respective salt and water. In step 4, SO2(RA) reacts with nascent oxgyen[O] to produce SO3. But SO3 is unstable. It immediately reacts with H2O in the step 5 to produce the corresponding acid, H2SO4. So we find that H2SO4 was used as a reactant and it is formed also in the product. So we add the five steps by multiplying suitable coefficients (3) to the step equations 4 and 5, to cancel 3[O] and 3SO3 from the two sides. K2O and Cr2O3 are also cancelled as such. On addition it gives an equation, which carries H2O and H2SO4 on either side. So we have to simplify further to get the balanced equation. We find that in the net balanced equation, H2SO4 has been wiped out from RHS.

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