Let's balance the given chemical equation step by step. The equation is:
[tex]\[ P_4(s) + NaOH(aq) + H_2O(l) \rightarrow PH_3(g) + Na_2HPO_3(aq) \][/tex]
### Step 1: Identify the elements involved.
The elements are:
- Phosphorus (P)
- Sodium (Na)
- Oxygen (O)
- Hydrogen (H)
### Step 2: Write down the number of atoms of each element on both sides.
- Phosphorus (P)
- Reactants: 4 (from [tex]\( P_4 \)[/tex])
- Products: 1 (from [tex]\( PH_3 \)[/tex]) + 2 (from [tex]\( Na_2HPO_3 \)[/tex])
- Hydrogen (H)
- Reactants: [tex]\( x \)[/tex] (from [tex]\( NaOH \)[/tex]) + [tex]\( y \)[/tex] (from [tex]\( H_2O \)[/tex])
- Products: 3 (from [tex]\( PH_3 \)[/tex]) + 4 (from [tex]\( Na_2HPO_3 \)[/tex])
- Sodium (Na)
- Reactants: [tex]\( z \)[/tex] (from [tex]\( NaOH \)[/tex])
- Products: 2 (from [tex]\( Na_2HPO_3 \)[/tex])
- Oxygen (O)
- Reactants: [tex]\( y \)[/tex] (from [tex]\( H_2O \)[/tex]) + [tex]\( z \)[/tex] (from [tex]\( NaOH \)[/tex])
- Products: 3 (from [tex]\( Na_2HPO_3 \)[/tex])
### Step 3: Balance the elements starting with phosphorus.
There are 4 phosphorus atoms in [tex]\( P_4 \)[/tex].
On the product side, each [tex]\( Na_2HPO_3 \)[/tex] contains 1 phosphorus atom, and each [tex]\( PH_3 \)[/tex] contains 1 phosphorus atom.
Since we want to balance the total number of phosphorus atoms, let's assume:
- [tex]\( x \)[/tex] moles of [tex]\( Na_2HPO_3 \)[/tex]
- [tex]\( y \)[/tex] moles of [tex]\( PH_3 \)[/tex]
Considering 4 phosphorus atoms on the reactant side:
[tex]\[ 4 = x + y \][/tex]
### Step 4: Balance sodium.
In [tex]\( Na_2HPO_3 \)[/tex], there are 2 sodium atoms, so:
[tex]\[ 2x \text{ (atoms of Na on product side from \( Na_2HPO_3 \))} \][/tex]
Since [tex]\( y = 4 - x \)[/tex], there is no sodium on the product side in [tex]\( PH_3 \)[/tex]:
[tex]\[ z \text{ (atoms of Na on reactant side from \( NaOH \))} = 2x \][/tex]
### Step 5: Balance hydrogen.
In [tex]\( PH_3 \)[/tex], there are 3 hydrogen atoms, and in each [tex]\( Na_2HPO_3 \)[/tex], there are 2 hydrogen atoms:
[tex]\[ 3y + 2x \text{ (atoms of H on product side)} = 6z + 2y \text{ (from \( NaOH \))} \][/tex]
### Step 6: Balance oxygen.
There is 1 oxygen atom in each molecule of [tex]\( Na_2HPO_3 \)[/tex]:
[tex]\[ x \text{ (oxygen atoms on product side)} = 2z (from \( NaOH \))\][/tex]
### Step 7: Solve the system of equations:
1. [tex]\( 4 = x + y \)[/tex]
2. [tex]\( z = 2x \)[/tex]
3. [tex]\( 6z + 2y = 2x + 3y + 6z + 2x \)[/tex]
Let's first balance [tex]\( y \)[/tex]:
Using [tex]\( y = 4 - x \)[/tex]:
2 [tex]\( ( number of sodium atoms ) \rightarrow 2y = 2(4 - x) = 8 - 2x \)[/tex]
So:
[tex]\[ 2x = 8 - 2x \][/tex]
This makes:
[tex]\[ z = 2x \][/tex]
\[ 6z + 3y = (4 \rightarrow x = 1 , 4-3)]\)
### Step 8: Conclusion:
The balanced chemical equation is:
\[ P_4 + 3H_2O + 4 NaOH \rightarrow 2Na_2HPO_3 + PH_3 ]
So, the balanced chemical equation using the smallest possible whole numbers is:
\[
P_4(s) + 4NaOH(aq) + 6H_2O(l) \rightarrow 2PH_3(g) + 2Na_2HPO_3(aq)
]
Therefore, the balanced equation with the smallest whole number coefficients is:
\[
P_4(s) + NaOH(aq) + H_2O(l) \rightarrow PH_3(g) + Na_2HPO_3(aq)
