Final answer:
To generate 7.65 moles of carbon tetrafluoride, 581.4 grams of fluorine are needed, based on the stoichiometry of the reaction and the molar mass of fluorine.
Step-by-step explanation:
To calculate the mass of fluorine needed to generate 7.65 moles of carbon tetrafluoride (CF4), we must first understand the stoichiometry of the reaction. The chemical formula of carbon tetrafluoride indicates that one mole of carbon is bonded to four moles of fluorine. Therefore, to produce one mole of CF4, four moles of fluorine are required.
To find the total amount of fluorine needed for 7.65 moles of CF4, we multiply the number of moles of CF4 by the four moles of fluorine per mole of CF4:
7.65 moles of CF4 × 4 moles of F per mole of CF4 = 30.6 moles of F
The molar mass of fluorine (F) is approximately 19 grams per mole. Therefore, to find the mass in grams:
30.6 moles of F × 19 grams/mole = 581.4 grams of F2
Thus, 581.4 grams of fluorine are required to produce 7.65 moles of carbon tetrafluoride.