Complete and balance the following equation:
NO2{-} + Cr2O7{2-} = Cr{3+} + NO3{-}
8H+(aq)+3NO2−(aq)+Cr2O72−(aq)→2Cr3+(aq)+3NO3−(aq)+4H2O(l)
Well, equations are all about balance, just like life. So here's the balanced equation:
3NO2{-} + Cr2O7{2-} = 2Cr{3+} + 3NO3{-}
See? It's all balanced now, just like a circus performer on a tightrope.
To complete and balance the equation, let's go step by step.
Step 1: Write down the unbalanced equation:
NO2{-} + Cr2O7{2-} = Cr{3+} + NO3{-}
Step 2: Balance the nitrogen atoms:
Since there is only one nitrogen atom on each side of the equation, the nitrogen atoms are already balanced.
Step 3: Balance the chromium atoms:
On the left side, we have 2 chromium atoms (Cr2O7{2-}), and on the right side, we have 1 chromium atom (Cr{3+}). To balance the chromium atoms, place a coefficient of 2 in front of Cr{3+}:
NO2{-} + Cr2O7{2-} = 2Cr{3+} + NO3{-}
Step 4: Balance the oxygen atoms:
On the left side, we have 7 oxygen atoms (Cr2O7{2-}), and on the right side, we have 3 oxygen atoms (NO3{-}). To balance the oxygen atoms, add a coefficient of 7 in front of NO3{-}:
NO2{-} + Cr2O7{2-} = 2Cr{3+} + 7NO3{-}
Step 5: Balance the charges:
On the left side, the overall charge is -1 (NO2{-}). On the right side, the overall charge is +6 (2Cr{3+}). To balance the charges, add a coefficient of 6 in front of NO2{-}:
6NO2{-} + Cr2O7{2-} = 2Cr{3+} + 7NO3{-}
Step 6: Final equation:
The balanced equation is: 6NO2{-} + Cr2O7{2-} = 2Cr{3+} + 7NO3{-}
To balance the given equation, we need to make sure that the number of each type of atom on both sides of the equation is equal. Let's break down the process step by step:
1. Start by counting the number of atoms of each element on both sides of the equation:
NO2(-) + Cr2O7(2-) = Cr(3+) + NO3(-)
On the left side, we have:
- 1 nitrogen (N)
- 3 oxygen (O)
- 1 chromium (Cr)
On the right side, we have:
- 1 nitrogen (N)
- 6 oxygen (O)
- 1 chromium (Cr)
2. Starting with the most complex polyatomic ion, balance the number of atoms one element at a time. In this case, let's balance the oxygen atoms first.
On the left side, we have 3 oxygen atoms from NO2(-) and 7 oxygen atoms from Cr2O7(2-), giving a total of 10 oxygen atoms.
On the right side, we have 3 oxygen atoms from NO3(-).
To balance the oxygen atoms, we need to multiply NO3(-) by 10/3 (or 3.33). This gives us:
NO2(-) + Cr2O7(2-) = Cr(3+) + (10/3)NO3(-)
3. Now that we have balanced the oxygen atoms, let's move on to balancing the nitrogen atoms.
On the left side, we have 1 nitrogen atom from NO2(-) and 2 nitrogen atoms from Cr2O7(2-), giving a total of 3 nitrogen atoms.
On the right side, we have 10/3 nitrogen atoms from NO3(-).
To balance the nitrogen atoms, we need to multiply NO2(-) by 10/9 (or 1.11) and Cr2O7(2-) by 10/9 (or 1.11). This gives us the balanced equation:
(10/9)NO2(-) + (10/9)Cr2O7(2-) = Cr(3+) + (10/3)NO3(-)
4. Finally, let's balance the chromium atoms.
On the left side, we have 2 chromium atoms from Cr2O7(2-).
On the right side, we have 1 chromium atom from Cr(3+).
To balance the chromium atoms, we need to multiply Cr(3+) by 2/1 (or 2). This gives us the final balanced equation:
(10/9)NO2(-) + (10/9)Cr2O7(2-) = 2Cr(3+) + (10/3)NO3(-)
And there you have it, the equation is complete and balanced!