Balance the chemical equation

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​+H2​O

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​+H2​O

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_2OH2​O 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​+3H2​O

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_2OH2​O).
  • 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​+3H2​O

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​+6H2​O

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​+6H2​O


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.

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