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Sagot :
Certainly! Let's analyze each of the given reactions step-by-step to determine if they make sense chemically. If they do, we'll also balance them.
### Reaction 1: [tex]\( Cl_{2} + NaOH \rightarrow NaCl + NaClO + H_{2}O \)[/tex]
This reaction appears to be the process of chlorine reacting with sodium hydroxide to produce sodium chloride, sodium hypochlorite, and water. This is known as a disproportionation reaction where chlorine is both oxidized and reduced.
To balance this reaction, let's start by counting the number of atoms of each element on both sides:
1. Chlorine (Cl): There are 2 chlorine atoms in [tex]\(Cl_{2}\)[/tex], and the products have 1 chlorine atom in [tex]\(NaCl\)[/tex] and 1 chlorine atom in [tex]\(NaClO\)[/tex], totaling 2 chlorine atoms.
2. Sodium (Na): There is 1 sodium atom in [tex]\(NaOH\)[/tex] and 2 sodium atoms in [tex]\(NaCl\)[/tex] and [tex]\(NaClO\)[/tex], totaling 2 sodium atoms.
3. Oxygen (O): There is 1 oxygen atom in [tex]\(NaOH\)[/tex] and 1 oxygen atom in [tex]\(NaClO\)[/tex] and 1 in [tex]\(H_{2}O\)[/tex], totaling 2 oxygen atoms.
4. Hydrogen (H): There is 1 hydrogen atom in [tex]\(NaOH\)[/tex] and 2 in [tex]\(H_{2}O\)[/tex], totaling 2 hydrogen atoms.
Balancing the equation:
[tex]\[ Cl_{2} + 2NaOH \rightarrow NaCl + NaClO + H_{2}O \][/tex]
### Reaction 2: [tex]\( I_{2} + KCl \rightarrow KI + Cl_{2} \)[/tex]
This reaction suggests that iodine is reacting with potassium chloride to produce potassium iodide and chlorine gas. However, iodine (I2) is less reactive than chlorine (Cl2) and typically cannot displace chlorine from potassium chloride to form potassium iodide.
Therefore, this reaction is not viable under standard conditions.
### Reaction 3: [tex]\( HNO_{3} + KCl \rightarrow KNO_{3} + HCl \)[/tex]
This is an acid-base reaction where nitric acid (HNO3) reacts with potassium chloride (KCl) to produce potassium nitrate (KNO3) and hydrochloric acid (HCl).
To balance this equation, let's count the atoms:
1. Hydrogen (H): 1 in [tex]\(HNO_{3}\)[/tex] and 1 in [tex]\(HCl\)[/tex]
2. Nitrogen (N): 1 in [tex]\(HNO_{3}\)[/tex] and 1 in [tex]\(KNO_{3}\)[/tex]
3. Oxygen (O): 3 in [tex]\(HNO_{3}\)[/tex] and 3 in [tex]\(KNO_{3}\)[/tex]
4. Potassium (K): 1 in [tex]\(KCl\)[/tex] and 1 in [tex]\(KNO_{3}\)[/tex]
5. Chlorine (Cl): 1 in [tex]\(KCl\)[/tex] and 1 in [tex]\(HCl\)[/tex]
The equation is already balanced:
[tex]\[ HNO_{3} + KCl \rightarrow KNO_{3} + HCl \][/tex]
### Reaction 4: [tex]\( FePO_{3} + RbCl \rightarrow Rb_{3}PO_{3} + FeCl_{3} \)[/tex]
We need to balance the equation for the number of each type of atom:
1. Iron (Fe): 1 on both sides.
2. Phosphorus (P): 1 on both sides.
3. Oxygen (O): 3 on both sides.
4. Rubidium (Rb): 1 in [tex]\(RbCl\)[/tex] and 3 in [tex]\(Rb_{3}PO_{3}\)[/tex]
5. Chlorine (Cl): 1 in [tex]\(RbCl\)[/tex] and 3 in [tex]\(FeCl_{3}\)[/tex]
Balancing the equation:
[tex]\[ FePO_{3} + 3RbCl \rightarrow Rb_{3}PO_{3} + FeCl_{3} \][/tex]
### Reaction 5: [tex]\( FeSO_{4} + Cu \rightarrow CuSO_{4} + Fe \)[/tex]
This is a single displacement reaction where copper displaces iron in iron(II) sulfate to produce copper(II) sulfate and iron metal.
Balancing the equation:
[tex]\[ FeSO_{4} + Cu \rightarrow CuSO_{4} + Fe \][/tex]
This equation is already balanced.
### Summary
Reactions 1, 3, 4, and 5 are feasible and can be balanced as described. However, reaction 2 is not feasible under standard conditions.
### Reaction 1: [tex]\( Cl_{2} + NaOH \rightarrow NaCl + NaClO + H_{2}O \)[/tex]
This reaction appears to be the process of chlorine reacting with sodium hydroxide to produce sodium chloride, sodium hypochlorite, and water. This is known as a disproportionation reaction where chlorine is both oxidized and reduced.
To balance this reaction, let's start by counting the number of atoms of each element on both sides:
1. Chlorine (Cl): There are 2 chlorine atoms in [tex]\(Cl_{2}\)[/tex], and the products have 1 chlorine atom in [tex]\(NaCl\)[/tex] and 1 chlorine atom in [tex]\(NaClO\)[/tex], totaling 2 chlorine atoms.
2. Sodium (Na): There is 1 sodium atom in [tex]\(NaOH\)[/tex] and 2 sodium atoms in [tex]\(NaCl\)[/tex] and [tex]\(NaClO\)[/tex], totaling 2 sodium atoms.
3. Oxygen (O): There is 1 oxygen atom in [tex]\(NaOH\)[/tex] and 1 oxygen atom in [tex]\(NaClO\)[/tex] and 1 in [tex]\(H_{2}O\)[/tex], totaling 2 oxygen atoms.
4. Hydrogen (H): There is 1 hydrogen atom in [tex]\(NaOH\)[/tex] and 2 in [tex]\(H_{2}O\)[/tex], totaling 2 hydrogen atoms.
Balancing the equation:
[tex]\[ Cl_{2} + 2NaOH \rightarrow NaCl + NaClO + H_{2}O \][/tex]
### Reaction 2: [tex]\( I_{2} + KCl \rightarrow KI + Cl_{2} \)[/tex]
This reaction suggests that iodine is reacting with potassium chloride to produce potassium iodide and chlorine gas. However, iodine (I2) is less reactive than chlorine (Cl2) and typically cannot displace chlorine from potassium chloride to form potassium iodide.
Therefore, this reaction is not viable under standard conditions.
### Reaction 3: [tex]\( HNO_{3} + KCl \rightarrow KNO_{3} + HCl \)[/tex]
This is an acid-base reaction where nitric acid (HNO3) reacts with potassium chloride (KCl) to produce potassium nitrate (KNO3) and hydrochloric acid (HCl).
To balance this equation, let's count the atoms:
1. Hydrogen (H): 1 in [tex]\(HNO_{3}\)[/tex] and 1 in [tex]\(HCl\)[/tex]
2. Nitrogen (N): 1 in [tex]\(HNO_{3}\)[/tex] and 1 in [tex]\(KNO_{3}\)[/tex]
3. Oxygen (O): 3 in [tex]\(HNO_{3}\)[/tex] and 3 in [tex]\(KNO_{3}\)[/tex]
4. Potassium (K): 1 in [tex]\(KCl\)[/tex] and 1 in [tex]\(KNO_{3}\)[/tex]
5. Chlorine (Cl): 1 in [tex]\(KCl\)[/tex] and 1 in [tex]\(HCl\)[/tex]
The equation is already balanced:
[tex]\[ HNO_{3} + KCl \rightarrow KNO_{3} + HCl \][/tex]
### Reaction 4: [tex]\( FePO_{3} + RbCl \rightarrow Rb_{3}PO_{3} + FeCl_{3} \)[/tex]
We need to balance the equation for the number of each type of atom:
1. Iron (Fe): 1 on both sides.
2. Phosphorus (P): 1 on both sides.
3. Oxygen (O): 3 on both sides.
4. Rubidium (Rb): 1 in [tex]\(RbCl\)[/tex] and 3 in [tex]\(Rb_{3}PO_{3}\)[/tex]
5. Chlorine (Cl): 1 in [tex]\(RbCl\)[/tex] and 3 in [tex]\(FeCl_{3}\)[/tex]
Balancing the equation:
[tex]\[ FePO_{3} + 3RbCl \rightarrow Rb_{3}PO_{3} + FeCl_{3} \][/tex]
### Reaction 5: [tex]\( FeSO_{4} + Cu \rightarrow CuSO_{4} + Fe \)[/tex]
This is a single displacement reaction where copper displaces iron in iron(II) sulfate to produce copper(II) sulfate and iron metal.
Balancing the equation:
[tex]\[ FeSO_{4} + Cu \rightarrow CuSO_{4} + Fe \][/tex]
This equation is already balanced.
### Summary
Reactions 1, 3, 4, and 5 are feasible and can be balanced as described. However, reaction 2 is not feasible under standard conditions.
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