Final answer:
To form 0.822 moles of fluorine gas, 37.51 grams of bromine trifluoride are required, calculated by first determining the moles of bromine trifluoride needed and then converting that amount into grams using its molar mass.
Step-by-step explanation:
To determine how many grams of bromine trifluoride are necessary to form 0.822 moles of fluorine gas, we must first write the balanced chemical equation for the reaction. For the sake of this example, let's assume the chemical equation is something like:
BrF3 → Br2 + 3F2
This shows that one mole of bromine trifluoride produces three moles of fluorine gas. Given that you want to produce 0.822 moles of fluorine, you would need 0.822 moles / 3 moles/mole = 0.274 moles of bromine trifluoride.
To find the mass of bromine trifluoride required, we need its molar mass. The molar mass of BrF3 is approximately 136.90 g/mol, so:
0.274 moles × 136.90 g/mol = 37.51 g
Therefore, 37.51 grams of bromine trifluoride are needed to produce 0.822 moles of fluorine gas.