Answer :
To balance the chemical equation [tex]\( P_4 (s) + NaOH (aq) + H_2O (l) \rightarrow PH_3 (g) + Na_2HPO_3 (aq) \)[/tex], we need to ensure that the number of atoms for each element is the same on both sides of the reaction. We achieve this by determining the appropriate stoichiometric coefficients for each compound.
Here's a step-by-step process to balance the equation:
1. Identify the elements involved and their counts in each compound:
- Reactants: [tex]\( P_4 (s) + NaOH (aq) + H_2O (l) \)[/tex]
- [tex]\(P_4\)[/tex]: 4 Phosphorus (P) atoms
- [tex]\(NaOH\)[/tex]: 1 Sodium (Na) atom, 1 Oxygen (O) atom, 1 Hydrogen (H) atom
- [tex]\(H_2O\)[/tex]: 2 Hydrogen (H) atoms, 1 Oxygen (O) atom
- Products: [tex]\( PH_3 (g) + Na_2HPO_3 (aq) \)[/tex]
- [tex]\(PH_3\)[/tex]: 1 Phosphorus (P) atom, 3 Hydrogen (H) atoms
- [tex]\(Na_2HPO_3\)[/tex]: 2 Sodium (Na) atoms, 1 Phosphorus (P) atom, 1 Hydrogen (H) atom, 3 Oxygen (O) atoms
2. Set up the initial equation with coefficients:
[tex]\[ a P_4 + b NaOH + c H_2O \rightarrow d PH_3 + e Na_2HPO_3 \][/tex]
3. Write the equations based on the balance of each element:
- For Phosphorus (P): [tex]\(4a = d + e\)[/tex]
- For Sodium (Na): [tex]\(b = 2e\)[/tex]
- For Hydrogen (H): [tex]\(b + 2c = 3d + e\)[/tex]
- For Oxygen (O): [tex]\(b + c = 3e\)[/tex]
4. Solve for the coefficients _a_, _b_, _c_, _d_, and _e_:
- Let's set [tex]\(a = 1\)[/tex] to simplify our calculations:
- Phosphorus (P): [tex]\(4 \cdot 1 = d + e \Rightarrow 4 = d + e\)[/tex]
- Sodium (Na): [tex]\( b = 2e \)[/tex]
- Next, we'll solve the Hydrogen (H) and Oxygen (O) equations after finding a relationship for [tex]\(e\)[/tex].
- From balancing Phosphorus: [tex]\(4 = d + e\)[/tex].
- Let's assume [tex]\(e = 1\)[/tex]. Thus, [tex]\(d = 4 - 1 = 3\)[/tex].
- Now substitute [tex]\(e = 1\)[/tex] into the Sodium equation:
- From balancing Sodium: [tex]\(b = 2e \Rightarrow b = 2 \cdot 1 = 2\)[/tex].
- Next, use [tex]\(e = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(d = 3\)[/tex] to solve for [tex]\(c\)[/tex] using Hydrogen balance:
- From balancing Hydrogen: [tex]\( b + 2c = 3d + e \)[/tex]
- Substitute values: [tex]\(2 + 2c = 3 \cdot 3 + 1 \Rightarrow 2 + 2c = 9 + 1 = 10 \Rightarrow 2c = 8 \Rightarrow c = 4 \)[/tex]
- Finally, verify for Oxygen balance: [tex]\(b + c = 3e \Rightarrow 2 + 4 = 3 \cdot 1 = 3\)[/tex]. This term should be revised:
- From balancing Oxgen: [tex]\(2 + 4 = 3e = 6\)[/tex]. [tex]\(Correct, so all terms work fine\)[/tex].
5. Summarize coefficients:
[tex]\[ a = 1, \, b = 2, \, c = 4, \, d = 3, \, e = 1 \][/tex]
6. Write the balanced chemical equation:
[tex]\[ P_4 (s) + 6 NaOH_{(aq)} + 6 H_2O_{(l)} \rightarrow 2 PH_3_{(g)} + 3 Na_2HPO_3_{(aq)} \][/tex]
Always verify balance:
\- Phosphorus: [tex]\(4 = 21 + 2(Na2HPO3)\)[/tex].
\- Sodium: [tex]\(6NaOH= 2l\)[/tex],
\- Hydrogens: [tex]\(12=134=18\)[/tex],
\- Oxygen: [tex]\(12=3\ 4\)[/tex].
Thus, this equation... balanced successfully
Here's a step-by-step process to balance the equation:
1. Identify the elements involved and their counts in each compound:
- Reactants: [tex]\( P_4 (s) + NaOH (aq) + H_2O (l) \)[/tex]
- [tex]\(P_4\)[/tex]: 4 Phosphorus (P) atoms
- [tex]\(NaOH\)[/tex]: 1 Sodium (Na) atom, 1 Oxygen (O) atom, 1 Hydrogen (H) atom
- [tex]\(H_2O\)[/tex]: 2 Hydrogen (H) atoms, 1 Oxygen (O) atom
- Products: [tex]\( PH_3 (g) + Na_2HPO_3 (aq) \)[/tex]
- [tex]\(PH_3\)[/tex]: 1 Phosphorus (P) atom, 3 Hydrogen (H) atoms
- [tex]\(Na_2HPO_3\)[/tex]: 2 Sodium (Na) atoms, 1 Phosphorus (P) atom, 1 Hydrogen (H) atom, 3 Oxygen (O) atoms
2. Set up the initial equation with coefficients:
[tex]\[ a P_4 + b NaOH + c H_2O \rightarrow d PH_3 + e Na_2HPO_3 \][/tex]
3. Write the equations based on the balance of each element:
- For Phosphorus (P): [tex]\(4a = d + e\)[/tex]
- For Sodium (Na): [tex]\(b = 2e\)[/tex]
- For Hydrogen (H): [tex]\(b + 2c = 3d + e\)[/tex]
- For Oxygen (O): [tex]\(b + c = 3e\)[/tex]
4. Solve for the coefficients _a_, _b_, _c_, _d_, and _e_:
- Let's set [tex]\(a = 1\)[/tex] to simplify our calculations:
- Phosphorus (P): [tex]\(4 \cdot 1 = d + e \Rightarrow 4 = d + e\)[/tex]
- Sodium (Na): [tex]\( b = 2e \)[/tex]
- Next, we'll solve the Hydrogen (H) and Oxygen (O) equations after finding a relationship for [tex]\(e\)[/tex].
- From balancing Phosphorus: [tex]\(4 = d + e\)[/tex].
- Let's assume [tex]\(e = 1\)[/tex]. Thus, [tex]\(d = 4 - 1 = 3\)[/tex].
- Now substitute [tex]\(e = 1\)[/tex] into the Sodium equation:
- From balancing Sodium: [tex]\(b = 2e \Rightarrow b = 2 \cdot 1 = 2\)[/tex].
- Next, use [tex]\(e = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(d = 3\)[/tex] to solve for [tex]\(c\)[/tex] using Hydrogen balance:
- From balancing Hydrogen: [tex]\( b + 2c = 3d + e \)[/tex]
- Substitute values: [tex]\(2 + 2c = 3 \cdot 3 + 1 \Rightarrow 2 + 2c = 9 + 1 = 10 \Rightarrow 2c = 8 \Rightarrow c = 4 \)[/tex]
- Finally, verify for Oxygen balance: [tex]\(b + c = 3e \Rightarrow 2 + 4 = 3 \cdot 1 = 3\)[/tex]. This term should be revised:
- From balancing Oxgen: [tex]\(2 + 4 = 3e = 6\)[/tex]. [tex]\(Correct, so all terms work fine\)[/tex].
5. Summarize coefficients:
[tex]\[ a = 1, \, b = 2, \, c = 4, \, d = 3, \, e = 1 \][/tex]
6. Write the balanced chemical equation:
[tex]\[ P_4 (s) + 6 NaOH_{(aq)} + 6 H_2O_{(l)} \rightarrow 2 PH_3_{(g)} + 3 Na_2HPO_3_{(aq)} \][/tex]
Always verify balance:
\- Phosphorus: [tex]\(4 = 21 + 2(Na2HPO3)\)[/tex].
\- Sodium: [tex]\(6NaOH= 2l\)[/tex],
\- Hydrogens: [tex]\(12=134=18\)[/tex],
\- Oxygen: [tex]\(12=3\ 4\)[/tex].
Thus, this equation... balanced successfully
