Consider the following reaction: Fe + Br₂ $\rightarrow$ FeBr₂
If 53.2 grams of iron react with an excess of bromine gas,what mass of FeBr₂ can form?
(FeBr₂ = 215.65 g/mol)

Question

Consider the following reaction: Fe + Br₂ $\rightarrow$ FeBr₂ If 53.2 grams of iron react with an excess of bromine gas,what mass of FeBr₂ can form? (FeBr₂ = 215.65 g/mol)

Quizplus · Accepted Answer

First, we need to correct the reaction and interpret it as Fe + Br2 → FeBr2. Given that the molar mass of FeBr2 is 215.65 g/mol, we calculate the moles of Fe (55.85 g/mol) used: 53.2 g / 55.85 g/mol = 0.952 moles. Since the reaction is 1:1, the same amount of moles of FeBr2 can form. Thus, the mass of FeBr2 formed is 0.952 moles * 215.65 g/mol = 205.3 g, which rounds to 205 g.

Consider the Following Reaction: Fe + Br2 →\rightarrow→ FeBr2 If 53

Consider the Following Reaction: Fe + Br₂ $\rightarrow$ FeBr₂
If 53