1. Write and balance the equation
2. Convert 3.75 g iron(III) nitrate to mols. mols = grams/molar mass
3. Using the coefficients in the balanced equation, convert mols iron(III) nitrate to mols of the product.
4. Finally, convert mols of the product to grams. g = mols x molar mass.
This 4-step procedure will work all simple (as opposed to limiting reagent problems) stoichiometry problems.
If 13.75 g of Iron (III) nitrate react with an excess of ammonium sulfide, how many grams of iron(III) sulfide are produced?
1 answer
1 answer