Answer:
The answer to part c is the amount of moles of FeBr₃ that will be formed is approximately 0.179 moles
Explanation:
The given chemical reaction is 2Fe + Br₂ → 2FeBr₃
The mass of iron, Fe, in the reaction = 10 g
a) The amount of moles of iron that reacted = The mass of iron in the reaction/(The molar mass of iron)
The atomic mass of iron = 56 amu
Therefore, one mole of iron = 56 grams
The amount of moles of iron that reacted = 10/(56) ≈ 0.179
The amount of moles of iron that reacted ≈ 0.179 moles
b) The amount of moles of bromine that reacted = 3/2 × The amount of moles of iron that reacted
∴ The amount of moles of bromine that reacted = 3/2 × 0.179 ≈ 0.2685
The amount of moles of bromine that reacted ≈ 0.2685 moles
c) The amount of moles of FeBr₃ that will be formed = The amount of moles of iron in the reaction ≈ 0.179 moles
d) The mass of one molecule of FeBr₃ = Mass of 1 atom of iron, Fe + 3 × The mass of 1 atom of bromine, Br
∴ The mass of one mole of FeBr₃ = 56 + 3 × 80 = 296
The mass of one mole of FeBr₃ = The molar mass of FeBr₃ = 296 grams
The mass of FeBr₃ formed = The molar mass of FeBr₃ × The number of moles of FeBr₃ formed
∴ The mass of FeBr₃ formed = 296 × 0.179 ≈ 52.984
The mass of FeBr₃ formed ≈ 52.984 grams
