Final answer:
0.125 moles of bromine are produced when fluorine gas reacts with 0.250 moles of potassium bromide.
Step-by-step explanation:
To determine the number of moles of bromine produced when fluorine gas reacts with 0.250 moles of potassium bromide, we need to balance the chemical equation and use stoichiometry. The balanced equation is:
3F2(g) + 6KBr(s) -> 3Br2(g) + 6KF(s)
From the equation, we can see that 6 moles of KBr react to produce 3 moles of Br2. Therefore, 0.250 moles of KBr will produce (0.250/6) x 3 = 0.125 moles of Br2.