Get the most out of your questions with the extensive resources available on IDNLearn.com. Get prompt and accurate answers to your questions from our community of knowledgeable experts.

Given the chemical reaction:

[tex]\[ C + 2F_2 \rightarrow CF_4 \][/tex]

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])?

Molar mass of [tex]\( CF_4 \)[/tex]: [tex]\( 88.01 \, \text{g/mol} \)[/tex]

[tex]\[ \text{[?] g CF}_4 \][/tex]


Sagot :

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]).
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.