Question
Chlorine can replace bromine in bromide compounds forming a chloride compound and elemental bromine. The following equation is an example of the reaction: 2KBr (aq) +Cl2 (aq) à 2KCl (aq) + Br2 (l)
a. When 1.855g of Cl2 and 3.205g of KBr are mixed in solution, which is the limiting reactant?
b. How many grams of KCl are formed?
a. When 1.855g of Cl2 and 3.205g of KBr are mixed in solution, which is the limiting reactant?
b. How many grams of KCl are formed?
Answers
GPT-4o mini
To determine the limiting reactant in the reaction \( 2KBr (aq) + Cl_2 (aq) \rightarrow 2KCl (aq) + Br_2 (l) \), we first need to convert the masses of \( Cl_2 \) and \( KBr \) into moles.
### a. Finding the Limiting Reactant
1. **Calculate the molar mass of the reactants:**
- **Chlorine (\( Cl_2 \)):**
\[
\text{Molar mass of } Cl_2 = 2 \times 35.453 \, \text{g/mol} = 70.906 \, \text{g/mol}
\]
- **Potassium bromide (\( KBr \)):**
\[
\text{Molar mass of } KBr = 39.098 \, \text{g/mol (K)} + 79.904 \, \text{g/mol (Br)} \approx 119.002 \, \text{g/mol}
\]
2. **Convert grams to moles:**
- **Moles of \( Cl_2 \):**
\[
\text{Moles of } Cl_2 = \frac{1.855 \, \text{g}}{70.906 \, \text{g/mol}} \approx 0.0262 \, \text{mol}
\]
- **Moles of \( KBr \):**
\[
\text{Moles of } KBr = \frac{3.205 \, \text{g}}{119.002 \, \text{g/mol}} \approx 0.0269 \, \text{mol}
\]
3. **Identify the stoichiometry of the reaction:**
From the balanced equation, we see that:
- 2 moles of \( KBr \) react with 1 mole of \( Cl_2 \).
4. **Determine the mole ratio needed:**
For every 1 mole of \( Cl_2 \), 2 moles of \( KBr \) are required. Therefore, to find out how many moles of \( KBr \) are required for the amount of \( Cl_2 \) we have:
Using the moles of \( Cl_2 \):
\[
\text{Moles of } KBr \text{ required} = 2 \times \text{Moles of } Cl_2 = 2 \times 0.0262 \approx 0.0524 \, \text{mol}
\]
5. **Compare available moles of \( KBr \):**
We have only \( 0.0269 \, \text{mol} \) of \( KBr \), which is less than \( 0.0524 \, \text{mol} \) needed. Thus, \( KBr \) is the limiting reactant.
### b. Calculate the mass of \( KCl \) formed
1. **Using the moles of the limiting reactant (KBr):**
From the balanced equation:
- 2 moles of \( KBr \) produce 2 moles of \( KCl \).
Thus, the moles of \( KCl \) produced will equal the moles of \( KBr \) used:
\[
\text{Moles of } KCl \text{ produced} = \text{Moles of } KBr = 0.0269 \, \text{mol}
\]
2. **Calculate the mass of \( KCl \):**
- **Molar mass of \( KCl \):**
\[
\text{Molar mass of } KCl = 39.098 \, \text{g/mol (K)} + 35.453 \, \text{g/mol (Cl)} = 74.551 \, \text{g/mol}
\]
- **Mass of \( KCl \):**
\[
\text{Mass of } KCl = \text{Moles of } KCl \times \text{Molar mass of } KCl = 0.0269 \, \text{mol} \times 74.551 \, \text{g/mol} \approx 2.005 \, \text{g}
\]
### Summary of Answers
a. The limiting reactant is **KBr**.
b. The mass of \( KCl \) formed is approximately **2.005 g**.
### a. Finding the Limiting Reactant
1. **Calculate the molar mass of the reactants:**
- **Chlorine (\( Cl_2 \)):**
\[
\text{Molar mass of } Cl_2 = 2 \times 35.453 \, \text{g/mol} = 70.906 \, \text{g/mol}
\]
- **Potassium bromide (\( KBr \)):**
\[
\text{Molar mass of } KBr = 39.098 \, \text{g/mol (K)} + 79.904 \, \text{g/mol (Br)} \approx 119.002 \, \text{g/mol}
\]
2. **Convert grams to moles:**
- **Moles of \( Cl_2 \):**
\[
\text{Moles of } Cl_2 = \frac{1.855 \, \text{g}}{70.906 \, \text{g/mol}} \approx 0.0262 \, \text{mol}
\]
- **Moles of \( KBr \):**
\[
\text{Moles of } KBr = \frac{3.205 \, \text{g}}{119.002 \, \text{g/mol}} \approx 0.0269 \, \text{mol}
\]
3. **Identify the stoichiometry of the reaction:**
From the balanced equation, we see that:
- 2 moles of \( KBr \) react with 1 mole of \( Cl_2 \).
4. **Determine the mole ratio needed:**
For every 1 mole of \( Cl_2 \), 2 moles of \( KBr \) are required. Therefore, to find out how many moles of \( KBr \) are required for the amount of \( Cl_2 \) we have:
Using the moles of \( Cl_2 \):
\[
\text{Moles of } KBr \text{ required} = 2 \times \text{Moles of } Cl_2 = 2 \times 0.0262 \approx 0.0524 \, \text{mol}
\]
5. **Compare available moles of \( KBr \):**
We have only \( 0.0269 \, \text{mol} \) of \( KBr \), which is less than \( 0.0524 \, \text{mol} \) needed. Thus, \( KBr \) is the limiting reactant.
### b. Calculate the mass of \( KCl \) formed
1. **Using the moles of the limiting reactant (KBr):**
From the balanced equation:
- 2 moles of \( KBr \) produce 2 moles of \( KCl \).
Thus, the moles of \( KCl \) produced will equal the moles of \( KBr \) used:
\[
\text{Moles of } KCl \text{ produced} = \text{Moles of } KBr = 0.0269 \, \text{mol}
\]
2. **Calculate the mass of \( KCl \):**
- **Molar mass of \( KCl \):**
\[
\text{Molar mass of } KCl = 39.098 \, \text{g/mol (K)} + 35.453 \, \text{g/mol (Cl)} = 74.551 \, \text{g/mol}
\]
- **Mass of \( KCl \):**
\[
\text{Mass of } KCl = \text{Moles of } KCl \times \text{Molar mass of } KCl = 0.0269 \, \text{mol} \times 74.551 \, \text{g/mol} \approx 2.005 \, \text{g}
\]
### Summary of Answers
a. The limiting reactant is **KBr**.
b. The mass of \( KCl \) formed is approximately **2.005 g**.