Consider the following reaction: Fe + Br₂ $\rightarrow$ FeBr₂
If 44.2 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 44.2 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, correct the chemical equation: Fe + Br2 → FeBr2. The molar mass of Fe is approximately 55.85 g/mol. Given 44.2 g of Fe, we calculate moles of Fe as 44.2 g / 55.85 g/mol ≈ 0.791 moles. The reaction is 1:1, so 0.791 moles of FeBr2 can form. The mass of FeBr2 is 0.791 moles × 215.65 g/mol ≈ 170.6 g, which rounds to 171 g.

Consider the Following Reaction: Fe + Br2 →\rightarrow→ FeBr2 If 44

Consider the Following Reaction: Fe + Br₂ $\rightarrow$ FeBr₂
If 44