Answer :
To solve this problem, we need to determine how much of the excess reagent remains after the reaction between chlorine gas ([tex]\(Cl_2\)[/tex]) and sodium fluoride ([tex]\(NaF\)[/tex]) takes place. Let's go through the steps:
1. Find the molar masses:
- Molar mass of [tex]\(Cl_2\)[/tex] is approximately 70.90 g/mol.
- Molar mass of [tex]\(NaF\)[/tex] is approximately 41.99 g/mol.
2. Calculate moles of each reactant:
- For [tex]\(Cl_2\)[/tex]:
[tex]\[
\text{Moles of } Cl_2 = \frac{37.8 \text{ g}}{70.90 \text{ g/mol}} \approx 0.533 \text{ moles}
\][/tex]
- For [tex]\(NaF\)[/tex]:
[tex]\[
\text{Moles of } NaF = \frac{39.4 \text{ g}}{41.99 \text{ g/mol}} \approx 0.938 \text{ moles}
\][/tex]
3. Determine the stoichiometry of the reaction:
- The reaction given is:
[tex]\[
Cl_2(g) + 2 \, NaF(aq) \rightarrow 2 \, NaCl(aq) + F_2(g)
\][/tex]
- This means 1 mole of [tex]\(Cl_2\)[/tex] reacts with 2 moles of [tex]\(NaF\)[/tex].
4. Find the limiting reagent:
- Calculate how much [tex]\(NaF\)[/tex] is required to react with all the [tex]\(Cl_2\)[/tex] available:
[tex]\[
\text{Required moles of } NaF = 0.533 \text{ moles of } Cl_2 \times 2 = 1.066 \text{ moles}
\][/tex]
- Since we only have 0.938 moles of [tex]\(NaF\)[/tex], [tex]\(NaF\)[/tex] is the limiting reagent.
5. Calculate the excess reagent ([tex]\(Cl_2\)[/tex]) left over:
- First, find how much [tex]\(Cl_2\)[/tex] will react with the available [tex]\(NaF\)[/tex]:
[tex]\[
\text{Moles of } Cl_2 \text{ used} = \frac{0.938 \text{ moles of } NaF}{2} = 0.469 \text{ moles}
\][/tex]
- Then calculate the moles of [tex]\(Cl_2\)[/tex] left:
[tex]\[
\text{Excess moles of } Cl_2 = 0.533 - 0.469 = 0.064 \text{ moles}
\][/tex]
- Convert this to grams:
[tex]\[
\text{Excess mass of } Cl_2 = 0.064 \text{ moles} \times 70.90 \text{ g/mol} \approx 4.54 \text{ g}
\][/tex]
The closest choice to our calculated result is 4.47 g, which matches the options given in the problem. Thus, the answer is 4.47 grams of [tex]\(Cl_2\)[/tex] are left over.
1. Find the molar masses:
- Molar mass of [tex]\(Cl_2\)[/tex] is approximately 70.90 g/mol.
- Molar mass of [tex]\(NaF\)[/tex] is approximately 41.99 g/mol.
2. Calculate moles of each reactant:
- For [tex]\(Cl_2\)[/tex]:
[tex]\[
\text{Moles of } Cl_2 = \frac{37.8 \text{ g}}{70.90 \text{ g/mol}} \approx 0.533 \text{ moles}
\][/tex]
- For [tex]\(NaF\)[/tex]:
[tex]\[
\text{Moles of } NaF = \frac{39.4 \text{ g}}{41.99 \text{ g/mol}} \approx 0.938 \text{ moles}
\][/tex]
3. Determine the stoichiometry of the reaction:
- The reaction given is:
[tex]\[
Cl_2(g) + 2 \, NaF(aq) \rightarrow 2 \, NaCl(aq) + F_2(g)
\][/tex]
- This means 1 mole of [tex]\(Cl_2\)[/tex] reacts with 2 moles of [tex]\(NaF\)[/tex].
4. Find the limiting reagent:
- Calculate how much [tex]\(NaF\)[/tex] is required to react with all the [tex]\(Cl_2\)[/tex] available:
[tex]\[
\text{Required moles of } NaF = 0.533 \text{ moles of } Cl_2 \times 2 = 1.066 \text{ moles}
\][/tex]
- Since we only have 0.938 moles of [tex]\(NaF\)[/tex], [tex]\(NaF\)[/tex] is the limiting reagent.
5. Calculate the excess reagent ([tex]\(Cl_2\)[/tex]) left over:
- First, find how much [tex]\(Cl_2\)[/tex] will react with the available [tex]\(NaF\)[/tex]:
[tex]\[
\text{Moles of } Cl_2 \text{ used} = \frac{0.938 \text{ moles of } NaF}{2} = 0.469 \text{ moles}
\][/tex]
- Then calculate the moles of [tex]\(Cl_2\)[/tex] left:
[tex]\[
\text{Excess moles of } Cl_2 = 0.533 - 0.469 = 0.064 \text{ moles}
\][/tex]
- Convert this to grams:
[tex]\[
\text{Excess mass of } Cl_2 = 0.064 \text{ moles} \times 70.90 \text{ g/mol} \approx 4.54 \text{ g}
\][/tex]
The closest choice to our calculated result is 4.47 g, which matches the options given in the problem. Thus, the answer is 4.47 grams of [tex]\(Cl_2\)[/tex] are left over.