posted by cambell .
iron (III) oxide is formed when iron combines with oxygen in the air. how many grams of Fe2O3 are formed when 16.7g of Fe reacts completely with oxygen? 4Fe(s) + 3O2(grams) => 2Fe2O3(s)
All of these stoichiometry problem are about the same. Remember these four steps.
Step 1. Write the balanced equation. You have that.
4Fe + 3O2 ==> 2Fe2O3
Step 2. Convert what you have (in this case g Fe) to mols remembering that mols = g/molar mass.
mols Fe = 16.7 g/55.8 = 0.299 mols Fe.
Step 3. Convert mols of what you have (in this case mols Fe) to mols of what you want (in this case mols Fe2O3) using the coefficients in the balanced equation from step 1.
mols Fe2O3 = mols Fe x (2 mols Fe2O3/4 mols Fe) = 0.299 x (2/4) = 0.299 x 1/2 = 0.150
Step 4. Now convert mols of Fe2O3 to grams usiong the equation from step 2 (but rearranged). There we had mols = g/molar mass. Rearrange to grams = mols x molar mass = 0.150 x molar mass Fe2O3 = xx g Fe2O3.
Check my work.
Post your work if you get stuck.