Answer :
Sure! Let's solve this problem step-by-step using the information provided.
### Step 1: Determine Moles of Ga and HF
- Gallium (Ga):
The molar mass of Ga is 69.72 g/mol. We have 250 grams of Ga.
[tex]\[
\text{Moles of Ga} = \frac{250 \, \text{g}}{69.72 \, \text{g/mol}} \approx 3.59 \, \text{moles}
\][/tex]
- Hydrofluoric Acid (HF):
The molar mass of HF is 20.04 g/mol. We have 314 grams of HF.
[tex]\[
\text{Moles of HF} = \frac{314 \, \text{g}}{20.04 \, \text{g/mol}} \approx 15.67 \, \text{moles}
\][/tex]
### Step 2: Reactant to Product Conversion
From the balanced equation, we know:
- 2 moles of Ga produces 2 moles of GaF₃.
- 6 moles of HF produces 2 moles of GaF₃.
- Ga to GaF₃:
Since it's a 1:1 ratio (2 Ga produces 2 GaF₃), the moles of GaF₃ formed from Ga is approximately 3.59 moles.
- HF to GaF₃:
According to the stoichiometry, 3 moles of HF forms 1 mole of GaF₃ (since 6 HF gives 2 GaF₃).
[tex]\[
\text{Moles of GaF}_3 \text{ from HF} = \frac{15.67 \, \text{moles HF}}{3} \approx 5.22 \, \text{moles}
\][/tex]
### Step 3: Identify the Limiting Reactant
The limiting reactant is the one that produces the least amount of product.
- Ga produces approximately 3.59 moles of GaF₃.
- HF could produce about 5.22 moles of GaF₃.
So, Ga is the limiting reactant because it produces less GaF₃.
### Step 4: Calculate Mass of GaF₃ Produced
The moles of GaF₃ that can be formed is limited by Ga: approximately 3.59 moles.
- Molar mass of GaF₃ is 126.72 g/mol.
[tex]\[
\text{Mass of GaF}_3 = 3.59 \, \text{moles} \times 126.72 \, \text{g/mol} \approx 454.39 \, \text{grams}
\][/tex]
### Step 5: Calculate Percent Yield
If 100 grams of GaF₃ were actually produced, we find the percent yield using the formula:
[tex]\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\][/tex]
[tex]\[
\text{Percent Yield} = \left( \frac{100 \, \text{g}}{454.39 \, \text{g}} \right) \times 100 \approx 22.01\%
\][/tex]
By following these steps, we conclude:
- Mass of GaF₃ produced with 250g of Ga: Approximately 454.39 grams.
- Mass of GaF₃ produced with 314g of HF: HF is not the limiting reactant, remains 454.39 grams.
- Limiting reactant: Ga.
- Percent yield: Approximately 22.01%.
### Step 1: Determine Moles of Ga and HF
- Gallium (Ga):
The molar mass of Ga is 69.72 g/mol. We have 250 grams of Ga.
[tex]\[
\text{Moles of Ga} = \frac{250 \, \text{g}}{69.72 \, \text{g/mol}} \approx 3.59 \, \text{moles}
\][/tex]
- Hydrofluoric Acid (HF):
The molar mass of HF is 20.04 g/mol. We have 314 grams of HF.
[tex]\[
\text{Moles of HF} = \frac{314 \, \text{g}}{20.04 \, \text{g/mol}} \approx 15.67 \, \text{moles}
\][/tex]
### Step 2: Reactant to Product Conversion
From the balanced equation, we know:
- 2 moles of Ga produces 2 moles of GaF₃.
- 6 moles of HF produces 2 moles of GaF₃.
- Ga to GaF₃:
Since it's a 1:1 ratio (2 Ga produces 2 GaF₃), the moles of GaF₃ formed from Ga is approximately 3.59 moles.
- HF to GaF₃:
According to the stoichiometry, 3 moles of HF forms 1 mole of GaF₃ (since 6 HF gives 2 GaF₃).
[tex]\[
\text{Moles of GaF}_3 \text{ from HF} = \frac{15.67 \, \text{moles HF}}{3} \approx 5.22 \, \text{moles}
\][/tex]
### Step 3: Identify the Limiting Reactant
The limiting reactant is the one that produces the least amount of product.
- Ga produces approximately 3.59 moles of GaF₃.
- HF could produce about 5.22 moles of GaF₃.
So, Ga is the limiting reactant because it produces less GaF₃.
### Step 4: Calculate Mass of GaF₃ Produced
The moles of GaF₃ that can be formed is limited by Ga: approximately 3.59 moles.
- Molar mass of GaF₃ is 126.72 g/mol.
[tex]\[
\text{Mass of GaF}_3 = 3.59 \, \text{moles} \times 126.72 \, \text{g/mol} \approx 454.39 \, \text{grams}
\][/tex]
### Step 5: Calculate Percent Yield
If 100 grams of GaF₃ were actually produced, we find the percent yield using the formula:
[tex]\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\][/tex]
[tex]\[
\text{Percent Yield} = \left( \frac{100 \, \text{g}}{454.39 \, \text{g}} \right) \times 100 \approx 22.01\%
\][/tex]
By following these steps, we conclude:
- Mass of GaF₃ produced with 250g of Ga: Approximately 454.39 grams.
- Mass of GaF₃ produced with 314g of HF: HF is not the limiting reactant, remains 454.39 grams.
- Limiting reactant: Ga.
- Percent yield: Approximately 22.01%.
