Before running your reaction, you determined that the mixture amount of product

Question

Before running your reaction, you determined that the mixture amount of product you could obtain was 3.74 g. However, after completeing your reaction the mass of product obtained was only 2.486 g. What mass is your theoretical yeild? What is your percent yeild?

Explanation / Answer

In order to determine the theoretical yield (maximum amount that could be formed) of iron (II) sulfide, you must first determine the limiting reagent/reactant. The limiting reactant is simply the reacting that present in a smaller amount (fewer number of moles not grams). Therefore you would need to calculated the number of moles of iron and sulfur that is present. You were given the mass of the reactants, you will need the molar mass of each in order to calculate the number of moles. The molar mass, simply tells you the number of grams of an element or compound in every one mole of that element or compound. This is determined by the atomic masses of each element (found on a periodic table). Calculating the number of moles: # moles of Iron (Fe) = (5.25 g) / (55.85 g/mol) = 0.0940 mol Fe # moles of sulfur (S) = (12.7 g) / (32.07 g/mol) = 0.396 mol S Because there are fewer moles of Iron than Sulfur, Iron is the limiting reactant. The amount of limiting reagent present determines the maximum amount of product that can be formed. To calculate this, you will have to use stoichiometry (using a balanced equation to relate the amounts of reactant(s) needed to react with the amount of product(s) formed). This reaction involves iron being combined with sulfur to form iron (II) sulfide. Therefore, the balanced chemical equation would be: Fe + S -> FeS. Therefore, for every one mole of iron reacting with one mole of sulfur, one mole of iron sulfide is formed. We now know that the maximum amount of iron 0.0940 mol. Therefore, we can determine the moles of product that can be formed as following: (0.0940 mol Fe) * (1 mol FeS / 1 mol Fe) = 0.0940 mol FeS **the problem was set up this way so that the mol Fe would cancel and mol FeS would remain** Once you have the moles of iron (II) sulfide, you can then use the molar mass to determine the mass (theoretical yield) of the product that is formed. The molar mass of the compound is simply calculated by using the atomic mass of all atoms involved. Molar mass of FeS = 55.85 g/mol + 32.07 g/mol = 87.92 g/mol Theoretical yield (or mass) of FeS = (0.0940 mol FeS) * (87.92g FeS/1 mol FeS) = 8.26 g FeS **I broke it up to show 87.92 g/mol as 87.92g FeS/1 mol FeS to show how the units would cancel**. Hope this helps!

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