Saved Bookmarks
| 1. |
Balance the following equation by oxidation number method: |
Answer» Solution :(i)` K_(2)Cr_(2)O_(7) + KCl+H_(2)SO_(4)rarrKHSO_(4)+CrO_(2)Cl_(2)+H_(2)O` Step 2 `K_(2)Cr_(2)O_(7)+KCl+KHSO_(4)+CrO_(2)Cl_(2)+H_(2)O` Step 3 to balance Cl atoms, KCl is multiplied by 4 `K_(2)Cr_(2)O_(7)+4KCl+H_(2)SO_(4)rarr2CrO_(2)Cl_(2)+KHSO_(4)+H_(2)O` Step 4. To balance K ATOM, `KHSO_(4)`is multiplied by 6. `K_(2)Cr_(2)O_(7)+4KC1+H_(2)SO_(4)rarr2CrO_(2)Cl_(2)+6KHSO_(4)+H_(2)O` Step 5. To balance O and H atoms, `H_(2)SO_(4)`is multiplied by 6, `H_(2)O` is multiplied by 3. `K_(2)Cr_(2)O_(7)+4KCl+6H_(2)SO_(4)rarr2CrO_(2)Cl_(2)+6KHSO_(4)+3H_(2)O` (ii) P+`HNO_(3)rarrH_(3)PO_(4)+NO_(2)+H_(2)O`  Step 2 P+`5HNO_(3)rarrH_(3)PO_(4)+5NO_(2)+H_(2)O` (iii) CuO+`NH_(3)rarrCu+N_(2)+H_(2)O`  Step 2.3CuO +`2NH_(3)rarr3Cu+N_(2)+H_(2)O` Step 3. To balance O and N, water is multiplied by 3. 3CuO+`2NH_(3)rarr3Cu+N_(2)+3H_(2)O` (iv) Zn+`HNO_(3)rarrZn(NO_(3))_(2)+NH_(4)NO_(3)+H_(2)O`  Step2. 4Zn+`HNO_(3)rarr4Zn(NO_(3))_(2)+NH_(4)NO_(3)+H_(2)O` Step 3. To balance N, `HNO_(3)` is multiplied by 10 4Zn+`10HNO_(3)rarr4Zn(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O` Step 4. To balance oxygen, `H_(2)O` is multiplied by 3 4Zn+`10HNO_(3)rarr4Zn(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O` |
|