To balance the chemical equation [tex]\( \text{H}_2(g) + \text{P}_4(s) \rightarrow \text{PH}_3(g) \)[/tex] using the smallest possible whole number coefficients, we follow a systematic approach. Here are the steps:
1. Write down the unbalanced equation:
[tex]\[ \text{H}_2 + \text{P}_4 \rightarrow \text{PH}_3 \][/tex]
2. Identify the number of atoms for each element in the reactants and products:
- Hydrogen (H):
- Reactants: 2 atoms (from H[tex]\(_2\)[/tex])
- Products: 3 atoms (from PH[tex]\(_3\)[/tex])
- Phosphorus (P):
- Reactants: 4 atoms (from P[tex]\(_4\)[/tex])
- Products: 1 atom (from PH[tex]\(_3\)[/tex])
3. Balance the phosphorus atoms first:
- There are 4 phosphorus atoms on the reactant side (P[tex]\(_4\)[/tex]).
- To balance, we need 4 phosphorus atoms on the product side.
[tex]\[ \text{P}_4 \rightarrow 4 \text{PH}_3 \][/tex]
- This gives us 4 phosphorus atoms and [tex]\( 4 \times 3 = 12 \)[/tex] hydrogen atoms on the product side.
4. Balance the hydrogen atoms:
- We now have 12 hydrogen atoms on the product side (from 4 PH[tex]\(_3\)[/tex]).
- We need 12 hydrogen atoms on the reactant side.
[tex]\[ 6 \text{H}_2 + \text{P}_4 \rightarrow 4 \text{PH}_3 \][/tex]
5. Check the balance:
- Hydrogen (H):
- Reactants: [tex]\(6 \times 2 = 12\)[/tex] atoms
- Products: [tex]\(4 \times 3 = 12\)[/tex] atoms
- Phosphorus (P):
- Reactants: 4 atoms
- Products: 4 atoms
6. The balanced equation is:
[tex]\[ 6 \text{H}_2 + \text{P}_4 \rightarrow 4 \text{PH}_3 \][/tex]
Therefore, the balanced chemical equation with the smallest possible whole number coefficients is:
[tex]\[ 6 \text{H}_2(g) + \text{P}_4(s) \rightarrow 4 \text{PH}_3(g) \][/tex]
