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To solve the problem of how many grams of carbon tetrafluoride ([tex]\(CF_4\)[/tex]) are produced from the complete reaction of 4.0 moles of fluorine ([tex]\(F_2\)[/tex]), let's go through each step methodically.
### Step 1: Understanding the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
[tex]\[ C + 2 F_2 \rightarrow CF_4 \][/tex]
This equation tells us that 1 mole of carbon ([tex]\(C\)[/tex]) reacts with 2 moles of fluorine ([tex]\(F_2\)[/tex]) to produce 1 mole of carbon tetrafluoride ([tex]\(CF_4\)[/tex]).
### Step 2: Determine the Moles of [tex]\(CF_4\)[/tex] Produced
Given data:
- Moles of [tex]\(F_2\)[/tex]: 4.0 moles
According to the balanced equation, 2 moles of [tex]\(F_2\)[/tex] produce 1 mole of [tex]\(CF_4\)[/tex]. Therefore, the moles of [tex]\(CF_4\)[/tex] produced can be calculated as follows:
[tex]\[ \text{Moles of } CF_4 \text{ produced} = \frac{\text{Moles of } F_2}{2} \][/tex]
[tex]\[ \text{Moles of } CF_4 \text{ produced} = \frac{4.0 \text{ moles of } F_2}{2} = 2.0 \text{ moles of } CF_4 \][/tex]
### Step 3: Calculate the Mass of [tex]\(CF_4\)[/tex] Produced
Given data:
- Molar mass of [tex]\(CF_4\)[/tex]: 88.01 g/mol
To find the mass of [tex]\(CF_4\)[/tex] produced, we use the number of moles of [tex]\(CF_4\)[/tex] and its molar mass. The formula for mass is:
[tex]\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \][/tex]
[tex]\[ \text{Mass of } CF_4 \text{ produced} = 2.0 \text{ moles of } CF_4 \times 88.01 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of } CF_4 \text{ produced} = 176.02 \text{ g} \][/tex]
### Conclusion
Thus, the complete reaction of 4.0 moles of fluorine ([tex]\(F_2\)[/tex]) produces 176.02 grams of carbon tetrafluoride ([tex]\(CF_4\)[/tex]).
### Step 1: Understanding the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
[tex]\[ C + 2 F_2 \rightarrow CF_4 \][/tex]
This equation tells us that 1 mole of carbon ([tex]\(C\)[/tex]) reacts with 2 moles of fluorine ([tex]\(F_2\)[/tex]) to produce 1 mole of carbon tetrafluoride ([tex]\(CF_4\)[/tex]).
### Step 2: Determine the Moles of [tex]\(CF_4\)[/tex] Produced
Given data:
- Moles of [tex]\(F_2\)[/tex]: 4.0 moles
According to the balanced equation, 2 moles of [tex]\(F_2\)[/tex] produce 1 mole of [tex]\(CF_4\)[/tex]. Therefore, the moles of [tex]\(CF_4\)[/tex] produced can be calculated as follows:
[tex]\[ \text{Moles of } CF_4 \text{ produced} = \frac{\text{Moles of } F_2}{2} \][/tex]
[tex]\[ \text{Moles of } CF_4 \text{ produced} = \frac{4.0 \text{ moles of } F_2}{2} = 2.0 \text{ moles of } CF_4 \][/tex]
### Step 3: Calculate the Mass of [tex]\(CF_4\)[/tex] Produced
Given data:
- Molar mass of [tex]\(CF_4\)[/tex]: 88.01 g/mol
To find the mass of [tex]\(CF_4\)[/tex] produced, we use the number of moles of [tex]\(CF_4\)[/tex] and its molar mass. The formula for mass is:
[tex]\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \][/tex]
[tex]\[ \text{Mass of } CF_4 \text{ produced} = 2.0 \text{ moles of } CF_4 \times 88.01 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of } CF_4 \text{ produced} = 176.02 \text{ g} \][/tex]
### Conclusion
Thus, the complete reaction of 4.0 moles of fluorine ([tex]\(F_2\)[/tex]) produces 176.02 grams of carbon tetrafluoride ([tex]\(CF_4\)[/tex]).
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