Question

Consider the following reaction: Fe + Br2 → FeBr2 If 113.18 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form? (Molar mass FeBr2 = 215.65 g/mol)

Answer 1

1) Balanced the chemical equation

$Fe+Br_2=FeBr_2$

2) Convert grams into moles

$\text{moles of Iron = }\frac{113.18\text{ grams of Iron}}{\square}\cdot\frac{1\text{ mole of Iron}}{55.8450\text{ grams of Iron}}=2.03\text{ moles of Iron}$

3) How many moles of FeBr2 can be formed?

$\text{moles of FeBr}_2=\frac{2.03\text{ moles of Fe}}{\square}\cdot\frac{1moleofFeBr_2}{1\text{ mole of Fe}}=2.03molesofFeBr_2$

4) Mass of FeBr2 produced.

$gramsofFeBr_2=\frac{2.03molesofFeBr_2}{\square}\cdot\frac{215.65gramsofFeBr_2}{1moleofFeBr_2}=437.77gramsofFeBr_2$

If 113.18 grams of Iron react with an excess of Bromine gas, 437.77 grams of FeBr2 will produce.