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Represent the following equation in ionic form K_(2)Cr_(2)O_(7)+7H_(2)SO_(4)+6FeSO_(4)=3Fe(SO_(4))_(3)+Cr_(2)(SO_(4))_(3)+7H_(2)O+K_(2)SO_(4) |
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Answer» SOLUTION :In this equation except, `H_(2)O`, all ionic in NATURE. Representing these compounds in ionic forms, `2K^(+)+Cr_(2)O_(7)^(2-)+14H^(+)+7SO_(4)^(2-)+6Fe^(2+)+6SO_(4)^(2-)to6Fe^(3+)+9SO_(4)^(2-)+2Cr^(3+)+3SO_(4)^(2-)+2K^(+)+SO_(4)^(2-)+7H_(2)O` `2K^(+)` ions and `13SO_(4)^(2-)` ions are common on both SIDES, so these are cancelled. The desired ionic equation reduces, to `Cr_(2)O_(7)^(2-)+14H^(+)+6Fe^(2+)=6Fe^(3+)+2Cr^(3+)+7H_(2)O` Total charge are equal on both sides, thus, the balanced ionic equation is the same as above. |
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