Answer :
Certainly! Let's tackle the problem step-by-step.
### Step 1: Calculate Moles of Each Reactant
First, determine the moles of each reactant given their masses and molar masses.
- Mass of [tex]\( CCl_4 \)[/tex] = 130.0 grams
- Molar mass of [tex]\( CCl_4 \)[/tex] = 153.823 g/mol
Moles of [tex]\( CCl_4 \)[/tex]:
[tex]\[ \text{Moles of } CCl_4 = \frac{\text{Mass of } CCl_4}{\text{Molar mass of } CCl_4} = \frac{130.0 \, \text{g}}{153.823 \, \text{g/mol}} = 0.8451271916423422 \, \text{moles} \][/tex]
- Mass of [tex]\( SbF_3 \)[/tex] = 95.0 grams
- Molar mass of [tex]\( SbF_3 \)[/tex] = 178.517 g/mol
Moles of [tex]\( SbF_3 \)[/tex]:
[tex]\[ \text{Moles of } SbF_3 = \frac{\text{Mass of } SbF_3}{\text{Molar mass of } SbF_3} = \frac{95.0 \, \text{g}}{178.517 \, \text{g/mol}} = 0.5321622030394865 \, \text{moles} \][/tex]
### Step 2: Determine the Limiting Reactant
Next, use the stoichiometric coefficients from the balanced equation to calculate the limiting reactant.
- Stoichiometric coefficients from the balanced equation:
- [tex]\( \text{CCl}_4: 3 \)[/tex]
- [tex]\( \text{SbF}_3: 2 \)[/tex]
Ratio of moles to coefficients for each reactant:
- For [tex]\( \text{CCl}_4 \)[/tex]:
[tex]\[ \text{Ratio of } CCl_4 = \frac{\text{Moles of } CCl_4}{\text{Coefficient of } CCl_4} = \frac{0.8451271916423422}{3} = 0.2817090638807807 \][/tex]
- For [tex]\( \text{SbF}_3 \)[/tex]:
[tex]\[ \text{Ratio of } SbF_3 = \frac{\text{Moles of } SbF_3}{\text{Coefficient of } SbF_3} = \frac{0.5321622030394865}{2} = 0.26608110151974323 \][/tex]
The limiting reactant is the one with the smaller ratio. In this case, [tex]\( \text{SbF}_3 \)[/tex] is the limiting reactant.
### Step 3: Calculate Moles of [tex]\( CCl_2F_2 \)[/tex] Produced
Using the limiting moles (which is the smaller ratio times the stoichiometric coefficient of the limiting reactant), we can determine the moles of [tex]\( CCl_2F_2 \)[/tex] produced.
[tex]\[ \text{Limiting moles} = 0.26608110151974323 \][/tex]
From the balanced chemical equation:
[tex]\[ 2 \text{ moles of } SbF_3 \rightarrow 3 \text{ moles of } CCl_2F_2 \][/tex]
Thus,
[tex]\[ \text{Moles of } CCl_2F_2 = 0.26608110151974323 \times \frac{3}{2} = 0.7982433045592296 \, \text{moles} \][/tex]
### Step 4: Calculate the Mass of [tex]\( CCl_2F_2 \)[/tex] Produced (i)
[tex]\[ \text{Molar mass of } CCl_2F_2 = 120.914 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } CCl_2F_2 = \text{Moles of } CCl_2F_2 \times \text{Molar mass of } CCl_2F_2 \][/tex]
[tex]\[ \text{Mass of } CCl_2F_2 = 0.7982433045592296 \times 120.914 = 96.5187909274747 \, \text{g} \][/tex]
So, the amount of [tex]\( CCl_2F_2 \)[/tex] obtained is approximately [tex]\( 96.52 \, \text{g} \)[/tex].
### Step 5: Calculate the Amount of Excess Reactant (ii)
Given that [tex]\( SbF_3 \)[/tex] is the limiting reactant, we can now calculate the excess amount of [tex]\( CCl_4 \)[/tex].
[tex]\[ \text{Moles of } CCl_4 \text{ used} = \text{Limiting moles} \times \text{Coefficient of } CCl_4 = 0.26608110151974323 \times 3 = 0.7982433045592296 \, \text{moles} \][/tex]
[tex]\[ \text{Moles of } CCl_4 \text{ initially} = 0.8451271916423422 \, \text{moles} \][/tex]
[tex]\[ \text{Moles of } CCl_4 \text{ remaining} = 0.8451271916423422 - 0.7982433045592296 = 0.0468838870831126 \, \text{moles} \][/tex]
[tex]\[ \text{Mass of excess } CCl_4 = \text{Moles of } CCl_4 \text{ remaining} \times \text{Molar mass of } CCl_4 \][/tex]
[tex]\[ \text{Mass of excess } CCl_4 = 0.0468838870831126 \times 153.823 = 7.211820162785619 \, \text{g} \][/tex]
So, the mass of the excess [tex]\( CCl_4 \)[/tex] is approximately [tex]\( 7.21 \, \text{g} \)[/tex].
### Summary
1. The amount of [tex]\( CCl_2F_2 \)[/tex] obtained is approximately [tex]\( 96.52 \, \text{grams} \)[/tex].
2. The amount of excess [tex]\( CCl_4 \)[/tex] remaining is approximately [tex]\( 7.21 \, \text{grams} \)[/tex].
### Step 1: Calculate Moles of Each Reactant
First, determine the moles of each reactant given their masses and molar masses.
- Mass of [tex]\( CCl_4 \)[/tex] = 130.0 grams
- Molar mass of [tex]\( CCl_4 \)[/tex] = 153.823 g/mol
Moles of [tex]\( CCl_4 \)[/tex]:
[tex]\[ \text{Moles of } CCl_4 = \frac{\text{Mass of } CCl_4}{\text{Molar mass of } CCl_4} = \frac{130.0 \, \text{g}}{153.823 \, \text{g/mol}} = 0.8451271916423422 \, \text{moles} \][/tex]
- Mass of [tex]\( SbF_3 \)[/tex] = 95.0 grams
- Molar mass of [tex]\( SbF_3 \)[/tex] = 178.517 g/mol
Moles of [tex]\( SbF_3 \)[/tex]:
[tex]\[ \text{Moles of } SbF_3 = \frac{\text{Mass of } SbF_3}{\text{Molar mass of } SbF_3} = \frac{95.0 \, \text{g}}{178.517 \, \text{g/mol}} = 0.5321622030394865 \, \text{moles} \][/tex]
### Step 2: Determine the Limiting Reactant
Next, use the stoichiometric coefficients from the balanced equation to calculate the limiting reactant.
- Stoichiometric coefficients from the balanced equation:
- [tex]\( \text{CCl}_4: 3 \)[/tex]
- [tex]\( \text{SbF}_3: 2 \)[/tex]
Ratio of moles to coefficients for each reactant:
- For [tex]\( \text{CCl}_4 \)[/tex]:
[tex]\[ \text{Ratio of } CCl_4 = \frac{\text{Moles of } CCl_4}{\text{Coefficient of } CCl_4} = \frac{0.8451271916423422}{3} = 0.2817090638807807 \][/tex]
- For [tex]\( \text{SbF}_3 \)[/tex]:
[tex]\[ \text{Ratio of } SbF_3 = \frac{\text{Moles of } SbF_3}{\text{Coefficient of } SbF_3} = \frac{0.5321622030394865}{2} = 0.26608110151974323 \][/tex]
The limiting reactant is the one with the smaller ratio. In this case, [tex]\( \text{SbF}_3 \)[/tex] is the limiting reactant.
### Step 3: Calculate Moles of [tex]\( CCl_2F_2 \)[/tex] Produced
Using the limiting moles (which is the smaller ratio times the stoichiometric coefficient of the limiting reactant), we can determine the moles of [tex]\( CCl_2F_2 \)[/tex] produced.
[tex]\[ \text{Limiting moles} = 0.26608110151974323 \][/tex]
From the balanced chemical equation:
[tex]\[ 2 \text{ moles of } SbF_3 \rightarrow 3 \text{ moles of } CCl_2F_2 \][/tex]
Thus,
[tex]\[ \text{Moles of } CCl_2F_2 = 0.26608110151974323 \times \frac{3}{2} = 0.7982433045592296 \, \text{moles} \][/tex]
### Step 4: Calculate the Mass of [tex]\( CCl_2F_2 \)[/tex] Produced (i)
[tex]\[ \text{Molar mass of } CCl_2F_2 = 120.914 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } CCl_2F_2 = \text{Moles of } CCl_2F_2 \times \text{Molar mass of } CCl_2F_2 \][/tex]
[tex]\[ \text{Mass of } CCl_2F_2 = 0.7982433045592296 \times 120.914 = 96.5187909274747 \, \text{g} \][/tex]
So, the amount of [tex]\( CCl_2F_2 \)[/tex] obtained is approximately [tex]\( 96.52 \, \text{g} \)[/tex].
### Step 5: Calculate the Amount of Excess Reactant (ii)
Given that [tex]\( SbF_3 \)[/tex] is the limiting reactant, we can now calculate the excess amount of [tex]\( CCl_4 \)[/tex].
[tex]\[ \text{Moles of } CCl_4 \text{ used} = \text{Limiting moles} \times \text{Coefficient of } CCl_4 = 0.26608110151974323 \times 3 = 0.7982433045592296 \, \text{moles} \][/tex]
[tex]\[ \text{Moles of } CCl_4 \text{ initially} = 0.8451271916423422 \, \text{moles} \][/tex]
[tex]\[ \text{Moles of } CCl_4 \text{ remaining} = 0.8451271916423422 - 0.7982433045592296 = 0.0468838870831126 \, \text{moles} \][/tex]
[tex]\[ \text{Mass of excess } CCl_4 = \text{Moles of } CCl_4 \text{ remaining} \times \text{Molar mass of } CCl_4 \][/tex]
[tex]\[ \text{Mass of excess } CCl_4 = 0.0468838870831126 \times 153.823 = 7.211820162785619 \, \text{g} \][/tex]
So, the mass of the excess [tex]\( CCl_4 \)[/tex] is approximately [tex]\( 7.21 \, \text{g} \)[/tex].
### Summary
1. The amount of [tex]\( CCl_2F_2 \)[/tex] obtained is approximately [tex]\( 96.52 \, \text{grams} \)[/tex].
2. The amount of excess [tex]\( CCl_4 \)[/tex] remaining is approximately [tex]\( 7.21 \, \text{grams} \)[/tex].
