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To balance the molecular equation for the reaction between the weak base sodium hypochlorite ([tex]\(\text{NaClO (aq)}\)[/tex]) and the strong acid sulfuric acid ([tex]\(\text{H}_2\text{SO}_4\text{(aq)}\)[/tex]), let's follow these steps:
1. List the reacting compounds and their respective elements:
- [tex]\(\text{NaClO}\)[/tex]
- Sodium (Na): 1
- Chlorine (Cl): 1
- Oxygen (O): 1
- [tex]\(\text{H}_2\text{SO}_4\)[/tex]
- Hydrogen (H): 2
- Sulfur (S): 1
- Oxygen (O): 4
2. Identify the products and balance each element:
- Products: [tex]\(\text{HClO} \text{ (aq)}\)[/tex] and [tex]\(\text{NaHSO}_4 \text{ (aq)}\)[/tex]
- [tex]\(\text{HClO}\)[/tex]
- Hydrogen (H): 1
- Chlorine (Cl): 1
- Oxygen (O): 1
- [tex]\(\text{NaHSO}_4\)[/tex]
- Sodium (Na): 1
- Hydrogen (H): 1
- Sulfur (S): 1
- Oxygen (O): 4
3. Ensure both sides of the reaction have equal numbers of atoms for each element:
- Left side (Reactants):
- Sodium (Na): 1 from [tex]\(\text{NaClO}\)[/tex]
- Chlorine (Cl): 1 from [tex]\(\text{NaClO}\)[/tex]
- Oxygen (O): 1 from [tex]\(\text{NaClO}\)[/tex] + 4 from [tex]\(\text{H}_2\text{SO}_4\)[/tex] = 5
- Hydrogen (H): 2 from [tex]\(\text{H}_2\text{SO}_4\)[/tex]
- Sulfur (S): 1 from [tex]\(\text{H}_2\text{SO}_4\)[/tex]
- Right side (Products):
- Sodium (Na): 1 from [tex]\(\text{NaHSO}_4\)[/tex]
- Chlorine (Cl): 1 from [tex]\(\text{HClO}\)[/tex]
- Oxygen (O): 1 from [tex]\(\text{HClO}\)[/tex] + 4 from [tex]\(\text{NaHSO}_4\)[/tex] = 5
- Hydrogen (H): 1 from [tex]\(\text{HClO}\)[/tex] + 1 from [tex]\(\text{NaHSO}_4\)[/tex] = 2
- Sulfur (S): 1 from [tex]\(\text{NaHSO}_4\)[/tex]
Since each element has the same number of atoms on both sides of the reaction equation, the equation is balanced.
4. Write the balanced equation:
[tex]\[ \text{NaClO (aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{HClO (aq)} + \text{NaHSO}_4\text{(aq)} \][/tex]
Thus, the balanced molecular reaction for the given weak base and strong acid is:
[tex]\[ \text{NaClO (aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{HClO (aq)} + \text{NaHSO}_4\text{(aq)} \][/tex]
1. List the reacting compounds and their respective elements:
- [tex]\(\text{NaClO}\)[/tex]
- Sodium (Na): 1
- Chlorine (Cl): 1
- Oxygen (O): 1
- [tex]\(\text{H}_2\text{SO}_4\)[/tex]
- Hydrogen (H): 2
- Sulfur (S): 1
- Oxygen (O): 4
2. Identify the products and balance each element:
- Products: [tex]\(\text{HClO} \text{ (aq)}\)[/tex] and [tex]\(\text{NaHSO}_4 \text{ (aq)}\)[/tex]
- [tex]\(\text{HClO}\)[/tex]
- Hydrogen (H): 1
- Chlorine (Cl): 1
- Oxygen (O): 1
- [tex]\(\text{NaHSO}_4\)[/tex]
- Sodium (Na): 1
- Hydrogen (H): 1
- Sulfur (S): 1
- Oxygen (O): 4
3. Ensure both sides of the reaction have equal numbers of atoms for each element:
- Left side (Reactants):
- Sodium (Na): 1 from [tex]\(\text{NaClO}\)[/tex]
- Chlorine (Cl): 1 from [tex]\(\text{NaClO}\)[/tex]
- Oxygen (O): 1 from [tex]\(\text{NaClO}\)[/tex] + 4 from [tex]\(\text{H}_2\text{SO}_4\)[/tex] = 5
- Hydrogen (H): 2 from [tex]\(\text{H}_2\text{SO}_4\)[/tex]
- Sulfur (S): 1 from [tex]\(\text{H}_2\text{SO}_4\)[/tex]
- Right side (Products):
- Sodium (Na): 1 from [tex]\(\text{NaHSO}_4\)[/tex]
- Chlorine (Cl): 1 from [tex]\(\text{HClO}\)[/tex]
- Oxygen (O): 1 from [tex]\(\text{HClO}\)[/tex] + 4 from [tex]\(\text{NaHSO}_4\)[/tex] = 5
- Hydrogen (H): 1 from [tex]\(\text{HClO}\)[/tex] + 1 from [tex]\(\text{NaHSO}_4\)[/tex] = 2
- Sulfur (S): 1 from [tex]\(\text{NaHSO}_4\)[/tex]
Since each element has the same number of atoms on both sides of the reaction equation, the equation is balanced.
4. Write the balanced equation:
[tex]\[ \text{NaClO (aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{HClO (aq)} + \text{NaHSO}_4\text{(aq)} \][/tex]
Thus, the balanced molecular reaction for the given weak base and strong acid is:
[tex]\[ \text{NaClO (aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{HClO (aq)} + \text{NaHSO}_4\text{(aq)} \][/tex]
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