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To solve the chemical equation [tex]\( NH_3 + NaClO \rightarrow N_2H_4 + NaCl + H_2O_2 \)[/tex] using stoichiometry, we need to balance the equation. Here is a detailed, step-by-step solution to achieve that:
1. Identify the number of atoms for each element on both sides:
- Reactants:
- [tex]\( NH_3 \)[/tex] has 1 nitrogen (N) and 3 hydrogens (H).
- [tex]\( NaClO \)[/tex] has 1 sodium (Na), 1 chlorine (Cl), and 1 oxygen (O).
- Products:
- [tex]\( N_2H_4 \)[/tex] has 2 nitrogens (N) and 4 hydrogens (H).
- [tex]\( NaCl \)[/tex] has 1 sodium (Na) and 1 chlorine (Cl).
- [tex]\( H_2O_2 \)[/tex] has 2 hydrogens (H) and 2 oxygens (O).
2. Set up the coefficients (unknowns) for each compound:
- Let [tex]\( a \)[/tex] be the coefficient for [tex]\( NH_3 \)[/tex].
- Let [tex]\( b \)[/tex] be the coefficient for [tex]\( NaClO \)[/tex].
- Let [tex]\( c \)[/tex] be the coefficient for [tex]\( N_2H_4 \)[/tex].
- Let [tex]\( d \)[/tex] be the coefficient for [tex]\( NaCl \)[/tex].
- Let [tex]\( e \)[/tex] be the coefficient for [tex]\( H_2O_2 \)[/tex].
3. Write equations based on the conservation of atoms:
- Nitrogen (N) atoms:
[tex]\( a = 2c \)[/tex]
- Hydrogen (H) atoms:
[tex]\( 3a = 4c + 2e \)[/tex]
- Sodium (Na) atoms:
[tex]\( b = d \)[/tex]
- Chlorine (Cl) atoms:
[tex]\( b = d \)[/tex]
- Oxygen (O) atoms:
[tex]\( b = 2e \)[/tex]
4. Solve the equations systematically:
From the nitrogen balance equation: [tex]\( a = 2c \)[/tex]
From the sodium and chlorine balances, we observe both are equal, so: [tex]\( b = d \)[/tex]
From the hydrogen balance: [tex]\( 3a = 4c + 2e \)[/tex]
Since [tex]\( a = 2c \)[/tex], substituting in the hydrogen equation:
[tex]\( 3(2c) = 4c + 2e \)[/tex]
[tex]\( 6c = 4c + 2e \)[/tex]
[tex]\( 2c = 2e \)[/tex]
[tex]\( c = e \)[/tex]
From the oxygen balance:
[tex]\( b = 2e \)[/tex]
Recall [tex]\( b = d \)[/tex], we write:
[tex]\( d = 2e \)[/tex]
5. Choose [tex]\( d \)[/tex] conveniently to ensure all coefficients are integers.
Let [tex]\( d = 1 \)[/tex], hence:
[tex]\[ b = 1 \][/tex]
[tex]\[ e = 0 \][/tex]
[tex]\[ c = 1/2 \][/tex]
[tex]\[ a = 1 \][/tex]
6. Therefore, the coefficients that will balance the equation are:
[tex]\[ \boxed{ 1NH_3 + 1NaClO → \frac{1}{2}N_2H_4 + 1NaCl + 0H_2O_2 } \][/tex]
However, fractions are not desirable in balanced equations, so multiply all coefficients by 2 to clear the fraction:
[tex]\[ 2NH_3 + 2NaClO → N_2H_4 + 2NaCl + 0H_2O_2 \][/tex]
Since we normally omit any zero coefficients, we get:
[tex]\[ 2NH_3 + 2NaClO → N_2H_4 + 2NaCl \][/tex]
The resulting balanced equation is:
[tex]\[ \boxed{ 2NH_3 + 2NaClO → N_2H_4 + 2NaCl } \][/tex]
1. Identify the number of atoms for each element on both sides:
- Reactants:
- [tex]\( NH_3 \)[/tex] has 1 nitrogen (N) and 3 hydrogens (H).
- [tex]\( NaClO \)[/tex] has 1 sodium (Na), 1 chlorine (Cl), and 1 oxygen (O).
- Products:
- [tex]\( N_2H_4 \)[/tex] has 2 nitrogens (N) and 4 hydrogens (H).
- [tex]\( NaCl \)[/tex] has 1 sodium (Na) and 1 chlorine (Cl).
- [tex]\( H_2O_2 \)[/tex] has 2 hydrogens (H) and 2 oxygens (O).
2. Set up the coefficients (unknowns) for each compound:
- Let [tex]\( a \)[/tex] be the coefficient for [tex]\( NH_3 \)[/tex].
- Let [tex]\( b \)[/tex] be the coefficient for [tex]\( NaClO \)[/tex].
- Let [tex]\( c \)[/tex] be the coefficient for [tex]\( N_2H_4 \)[/tex].
- Let [tex]\( d \)[/tex] be the coefficient for [tex]\( NaCl \)[/tex].
- Let [tex]\( e \)[/tex] be the coefficient for [tex]\( H_2O_2 \)[/tex].
3. Write equations based on the conservation of atoms:
- Nitrogen (N) atoms:
[tex]\( a = 2c \)[/tex]
- Hydrogen (H) atoms:
[tex]\( 3a = 4c + 2e \)[/tex]
- Sodium (Na) atoms:
[tex]\( b = d \)[/tex]
- Chlorine (Cl) atoms:
[tex]\( b = d \)[/tex]
- Oxygen (O) atoms:
[tex]\( b = 2e \)[/tex]
4. Solve the equations systematically:
From the nitrogen balance equation: [tex]\( a = 2c \)[/tex]
From the sodium and chlorine balances, we observe both are equal, so: [tex]\( b = d \)[/tex]
From the hydrogen balance: [tex]\( 3a = 4c + 2e \)[/tex]
Since [tex]\( a = 2c \)[/tex], substituting in the hydrogen equation:
[tex]\( 3(2c) = 4c + 2e \)[/tex]
[tex]\( 6c = 4c + 2e \)[/tex]
[tex]\( 2c = 2e \)[/tex]
[tex]\( c = e \)[/tex]
From the oxygen balance:
[tex]\( b = 2e \)[/tex]
Recall [tex]\( b = d \)[/tex], we write:
[tex]\( d = 2e \)[/tex]
5. Choose [tex]\( d \)[/tex] conveniently to ensure all coefficients are integers.
Let [tex]\( d = 1 \)[/tex], hence:
[tex]\[ b = 1 \][/tex]
[tex]\[ e = 0 \][/tex]
[tex]\[ c = 1/2 \][/tex]
[tex]\[ a = 1 \][/tex]
6. Therefore, the coefficients that will balance the equation are:
[tex]\[ \boxed{ 1NH_3 + 1NaClO → \frac{1}{2}N_2H_4 + 1NaCl + 0H_2O_2 } \][/tex]
However, fractions are not desirable in balanced equations, so multiply all coefficients by 2 to clear the fraction:
[tex]\[ 2NH_3 + 2NaClO → N_2H_4 + 2NaCl + 0H_2O_2 \][/tex]
Since we normally omit any zero coefficients, we get:
[tex]\[ 2NH_3 + 2NaClO → N_2H_4 + 2NaCl \][/tex]
The resulting balanced equation is:
[tex]\[ \boxed{ 2NH_3 + 2NaClO → N_2H_4 + 2NaCl } \][/tex]
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