The complete balanced equation is $ {{H}_{2}}S+{{K}_{2}}Cr{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to $ $ {{H}_{2}}S+{{K}_{2}}Cr{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to $ (A) $ 3{{H}_{2}}S+2{{K}_{2}}Cr{{O}_{4}}+5{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+2{{K}_{2}}S{{O}_{4}}+8{{H}_{2}}O+3S $ (B) $ 3{{H}_{2}}S+{{K}_{2}}Cr{{O}_{4}}+5{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+8{{H}_{2}}O+3S $ (C) $ {{H}_{2}}S+2{{K}_{2}}Cr{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+2{{K}_{2}}S{{O}_{4}}+8{{H}_{2}}O+3S $ (D) None of these

Question

The complete balanced equation is  $  {{H}_{2}}S+{{K}_{2}}Cr{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to   $     $  {{H}_{2}}S+{{K}_{2}}Cr{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to   $  (A)   $  3{{H}_{2}}S+2{{K}_{2}}Cr{{O}_{4}}+5{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+2{{K}_{2}}S{{O}_{4}}+8{{H}_{2}}O+3S  $  (B)   $  3{{H}_{2}}S+{{K}_{2}}Cr{{O}_{4}}+5{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+8{{H}_{2}}O+3S  $  (C)   $  {{H}_{2}}S+2{{K}_{2}}Cr{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+2{{K}_{2}}S{{O}_{4}}+8{{H}_{2}}O+3S  $  (D) None of these

Vedantu Content Team · Accepted Answer

Hint :First write the possible product can be formed from the given reactants. The possible product is  $ {{H}_{2}}S+{{K}_{2}}Cr{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+S $ . Calculate the number of Cr, K, S, on both sides to write the balance equation. First try to balance all the elements other than oxygen and hydrogen. Finally, balance the oxygen and hydrogen.Complete Step By Step Answer:First, write down the possible product can be formed from the given reactants $ {{H}_{2}}S+{{K}_{2}}Cr{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+S $ Now, balance the elements Cr, K, SThe oxidation number of Cr on the  $ {{K}_{2}}Cr{{O}_{4}} $ reactant side is  $ +6 $ The oxidation number of Cr on the  $ C{{r}_{2}}{{(S{{O}_{4}})}_{3}} $ product side is  $ +3 $ The oxidation number of Cr was reduced from  $ +6 $  to  $ +3 $ The change in oxidation number of Cr is  $ +3 $ Therefore, for  $ 2Cr $ is  $ +6 $ Hence, balanced The oxidation number of S is changed from  $ -2 $  to  $ 0 $ The increase in oxidation number is balanced with decrease in oxidation number by multiplying  $ {{H}_{2}}S $  and $ S $ with  $ +3 $ Now balancing the equation, we get $ 3{{H}_{2}}S+2{{K}_{2}}Cr{{O}_{4}}+5{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+2{{K}_{2}}S{{O}_{4}}+3S $ Now we should balance the oxygen atomsOxygen atoms are balanced by adding eight water molecules on the product sideSo, now the equation becomes in such a way that $ 3{{H}_{2}}S+2{{K}_{2}}Cr{{O}_{4}}+5{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+2{{K}_{2}}S{{O}_{4}}+3S+8{{H}_{2}}O $ Therefore, the first option is the complete balanced equationThe number of Cr is  $ 2 $  on both sides in the balanced equation.The number of K is  $ 4 $ on both sides in the balanced equation.The number of S is  $ 8 $  on both sides in the balanced equation.The number of O is  $ 28 $  on both sides in the balanced equation.The number of H is  $ 16 $  on both sides in the balanced equation.The correct option is option A.Note :  $  3{{H}_{2}}S+2{{K}_{2}}Cr{{O}_{4}}+5{{H}_{2}}S{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+2{{K}_{2}}S{{O}_{4}}+3S+8{{H}_{2}}O  $  Therefore, the first option is the complete balanced equationThe correct option is option A.

The complete balanced equation is H2S+K2CrO4+H2SO4→ H2S+K2CrO4+H2SO4→ (A) 3H2S+2K2CrO4+5H2SO4→Cr2(SO4)3+2K2SO4+8H2O+3S (B) 3H2S+K2CrO4+5H2SO4→Cr2(SO4)3+K2SO4+8H2O+3S (C) H2S+2K2CrO4+3H2SO4→Cr2(SO4)3+2K2SO4+8H2O+3S (D) None of these

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