Balance the chemical equation: NH3 + O2 -> N2 +H2O
The Correct Answer and Explanation is:
Let’s balance the chemical equation:
Unbalanced equation:
NH3+O2→N2+H2ONH_3 + O_2 \rightarrow N_2 + H_2ONH3+O2→N2+H2O
Step 1: Balance the Nitrogen (N) atoms
- On the left side, there is 1 nitrogen atom in NH3NH_3NH3 and on the right side, there are 2 nitrogen atoms in N2N_2N2.
- To balance nitrogen, place a coefficient of 2 in front of NH3NH_3NH3:
2NH3+O2→N2+H2O2NH_3 + O_2 \rightarrow N_2 + H_2O2NH3+O2→N2+H2O
Now, there are 2 nitrogen atoms on both sides.
Step 2: Balance the Hydrogen (H) atoms
- On the left side, there are 6 hydrogen atoms from 2 molecules of NH3NH_3NH3 (since each NH3NH_3NH3 has 3 hydrogen atoms).
- On the right side, each H2OH_2OH2O molecule contains 2 hydrogen atoms. So, we need 3 molecules of water to have 6 hydrogen atoms.
2NH3+O2→N2+3H2O2NH_3 + O_2 \rightarrow N_2 + 3H_2O2NH3+O2→N2+3H2O
Now, the hydrogen atoms are balanced.
Step 3: Balance the Oxygen (O) atoms
- On the right side, there are 3 oxygen atoms (from 3 molecules of H2OH_2OH2O).
- On the left side, oxygen comes from O2O_2O2, which means we need 1.5 molecules of O2O_2O2 to supply 3 oxygen atoms.
2NH3+1.5O2→N2+3H2O2NH_3 + 1.5O_2 \rightarrow N_2 + 3H_2O2NH3+1.5O2→N2+3H2O
Step 4: Eliminate the fraction
- To eliminate the fraction, multiply the entire equation by 2:
4NH3+3O2→2N2+6H2O4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O4NH3+3O2→2N2+6H2O
Now the equation is balanced, with 4 nitrogen atoms, 12 hydrogen atoms, and 6 oxygen atoms on both sides.
Final balanced equation:
4NH3+3O2→2N2+6H2O4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O4NH3+3O2→2N2+6H2O
Explanation:
Balancing chemical equations involves ensuring that the number of atoms for each element is the same on both sides. This is based on the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. In this case, we adjusted the coefficients in front of each molecule to match the number of atoms of nitrogen, hydrogen, and oxygen on both sides of the equation. This process ensures that the equation is stoichiometrically correct.
