(NH4)2 CR2 O7 –> N2+ CR2O3 + H2O
The reaction you provided is the thermal decomposition of ammonium dichromate \((NH_4)_2Cr_2O_7\). When heated, it decomposes to form nitrogen gas \((N_2)\), chromium(III) oxide \((Cr_2O_3)\), and water \((H_2O)\).
To balance the reaction, we can start by writing the unbalanced equation:
\[ (NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + H_2O \]
Balancing the Equation:
Count the number of each element on both sides of the equation.
On the left (reactants):
- Nitrogen (N): 2 (from \((NH_4)_2\))
- Hydrogen (H): 8 (from \((NH_4)_2\))
- Chromium (Cr): 2 (from \(Cr_2O_7\))
- Oxygen (O): 7 (from \(Cr_2O_7\))
On the right (products):
- Nitrogen (N): 2 (from \(N_2\))
- Hydrogen (H): 2 (from \(H_2O\) - to be balanced)
- Chromium (Cr): 2 (from \(Cr_2O_3\))
- Oxygen (O): 3 (from \(Cr_2O_3\)) + 1 (from \(H_2O\)) = 4
Balance the Hydrogen: Since there are 8 hydrogens on the left, we need 4 water molecules on the right to balance hydrogen: \[ (NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O \]
Now recap the counts on the right:
- N: 2 (from \(N_2\))
- H: 8 (from \(4H_2O\))
- Cr: 2 (from \(Cr_2O_3\))
- O: 3 (from \(Cr_2O_3\)) + 4 (from \(4H_2O\)) = 7
The balanced chemical reaction is then:
\[ (NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O \]
This equation shows that one mole of ammonium dichromate decomposes to yield one mole of nitrogen gas, one mole of chromium(III) oxide, and four moles of water.
