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Sagot :
Certainly! Let's balance each of the equations step-by-step.
### Equation a: [tex]\( NH_3 + Cl_2 \rightarrow A + N_2 \)[/tex]
1. Identify the atoms involved:
- Left side: 1 N, 3 H, 2 Cl
- Right side: 2 N (from [tex]\(N_2\)[/tex])
2. Balancing Nitrogen:
- For balancing Nitrogen on the right, there are 2 nitrogen atoms in [tex]\(N_2\)[/tex]. Therefore, on the left, we need to adjust to have 2 nitrogen atoms: [tex]\(2 NH_3\)[/tex].
Now we have:
[tex]\[ 2 NH_3 + Cl_2 \rightarrow A + N_2 \][/tex]
- Left side: 2 N, 6 H, 2 Cl
3. Identify product A:
- A suitable product containing [tex]\( NH _4^+ \)[/tex] and [tex]\( Cl^- \)[/tex] from the left side atoms would be [tex]\( NH_4Cl \)[/tex].
4. Balancing the equation:
[tex]\[ 2 NH_3 + 3 Cl_2 \rightarrow 2 NH_4Cl + N_2 \][/tex]
Now the left atoms match the right.
Thus, [tex]\(A = 2 NH_4Cl \)[/tex].
### Equation b: [tex]\( CaOCl_2 + NH_3 \rightarrow B + CaCl_2 + 3 H_2O \)[/tex]
1. Identify the atoms involved:
- Left side: 1 Ca, 1 O, 2 Cl, 1 N, 3 H
- Right side: 1 Ca, 2 Cl, 3 H2O (6 H, 3 O)
2. Balancing the nitrogen and hydrogens:
- The presence of [tex]\(CaCl_2\)[/tex] and 6 H in the form of [tex]\(3 H_2O\)[/tex] indicates that compound B must contain nitrogen, hydrogen, and chlorine.
3. Balancing chlorine factor in B:
- [tex]\( NH_4Cl\)[/tex] seems appropriate given the remaining.
4. Balancing the equation:
[tex]\[ CaOCl_2 + 2 NH_3 \rightarrow 2 NH_4Cl + CaCl_2 + H_2O \][/tex]
This ensures every atom has equal representation.
Thus, [tex]\(B = 2 NH_4Cl\)[/tex].
### Equation c: [tex]\( NaOBr + NH_3 \rightarrow N_2 + C + 3 H_2O \)[/tex]
1. Identify atoms involved:
- Left side: 1 Na, 1 O, 1 Br, 1 N, 3 H
- Right side: 2 N (from [tex]\(N_2\)[/tex]), 6 H (from [tex]\(3 H_2O\)[/tex])
2. Balancing and examination of remaining on new compound:
- Considering remaining elements as NaOH and observed.
3. Balancing the equation:
[tex]\[ NaOBr + NH_3 \rightarrow NaOH + NaBr + N_2 + 3 H_2O \][/tex]
Thus, [tex]\( C = NaOH + NaBr \)[/tex].
### Equation d: [tex]\( P_4O_{10} + D \rightarrow P_4 + CO \)[/tex]
1. Identify atoms involved:
- Left side: 4 P, 10 O
- Right side: 4 P, requiring proper oxygen balancing [tex]\(CO\)[/tex] product.
2. Balancing remaining oxides:
- [tex]\(5 CO\)[/tex] needed to balance 10 O atoms for each 5 component C interaction.
3. Balancing the equation:
[tex]\[ P_4O_{10} + 5 C \rightarrow P_4 + 5 CO \][/tex]
Thus, [tex]\( D = 5 C \)[/tex].
In conclusion:
- [tex]\(A = 2 NH_4Cl\)[/tex]
- [tex]\(B = 2 NH_4Cl\)[/tex]
- [tex]\(C = NaOH + NaBr\)[/tex]
- [tex]\(D = 5 C\)[/tex]
### Equation a: [tex]\( NH_3 + Cl_2 \rightarrow A + N_2 \)[/tex]
1. Identify the atoms involved:
- Left side: 1 N, 3 H, 2 Cl
- Right side: 2 N (from [tex]\(N_2\)[/tex])
2. Balancing Nitrogen:
- For balancing Nitrogen on the right, there are 2 nitrogen atoms in [tex]\(N_2\)[/tex]. Therefore, on the left, we need to adjust to have 2 nitrogen atoms: [tex]\(2 NH_3\)[/tex].
Now we have:
[tex]\[ 2 NH_3 + Cl_2 \rightarrow A + N_2 \][/tex]
- Left side: 2 N, 6 H, 2 Cl
3. Identify product A:
- A suitable product containing [tex]\( NH _4^+ \)[/tex] and [tex]\( Cl^- \)[/tex] from the left side atoms would be [tex]\( NH_4Cl \)[/tex].
4. Balancing the equation:
[tex]\[ 2 NH_3 + 3 Cl_2 \rightarrow 2 NH_4Cl + N_2 \][/tex]
Now the left atoms match the right.
Thus, [tex]\(A = 2 NH_4Cl \)[/tex].
### Equation b: [tex]\( CaOCl_2 + NH_3 \rightarrow B + CaCl_2 + 3 H_2O \)[/tex]
1. Identify the atoms involved:
- Left side: 1 Ca, 1 O, 2 Cl, 1 N, 3 H
- Right side: 1 Ca, 2 Cl, 3 H2O (6 H, 3 O)
2. Balancing the nitrogen and hydrogens:
- The presence of [tex]\(CaCl_2\)[/tex] and 6 H in the form of [tex]\(3 H_2O\)[/tex] indicates that compound B must contain nitrogen, hydrogen, and chlorine.
3. Balancing chlorine factor in B:
- [tex]\( NH_4Cl\)[/tex] seems appropriate given the remaining.
4. Balancing the equation:
[tex]\[ CaOCl_2 + 2 NH_3 \rightarrow 2 NH_4Cl + CaCl_2 + H_2O \][/tex]
This ensures every atom has equal representation.
Thus, [tex]\(B = 2 NH_4Cl\)[/tex].
### Equation c: [tex]\( NaOBr + NH_3 \rightarrow N_2 + C + 3 H_2O \)[/tex]
1. Identify atoms involved:
- Left side: 1 Na, 1 O, 1 Br, 1 N, 3 H
- Right side: 2 N (from [tex]\(N_2\)[/tex]), 6 H (from [tex]\(3 H_2O\)[/tex])
2. Balancing and examination of remaining on new compound:
- Considering remaining elements as NaOH and observed.
3. Balancing the equation:
[tex]\[ NaOBr + NH_3 \rightarrow NaOH + NaBr + N_2 + 3 H_2O \][/tex]
Thus, [tex]\( C = NaOH + NaBr \)[/tex].
### Equation d: [tex]\( P_4O_{10} + D \rightarrow P_4 + CO \)[/tex]
1. Identify atoms involved:
- Left side: 4 P, 10 O
- Right side: 4 P, requiring proper oxygen balancing [tex]\(CO\)[/tex] product.
2. Balancing remaining oxides:
- [tex]\(5 CO\)[/tex] needed to balance 10 O atoms for each 5 component C interaction.
3. Balancing the equation:
[tex]\[ P_4O_{10} + 5 C \rightarrow P_4 + 5 CO \][/tex]
Thus, [tex]\( D = 5 C \)[/tex].
In conclusion:
- [tex]\(A = 2 NH_4Cl\)[/tex]
- [tex]\(B = 2 NH_4Cl\)[/tex]
- [tex]\(C = NaOH + NaBr\)[/tex]
- [tex]\(D = 5 C\)[/tex]
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