Question

A student has 20 grams of iron and 40 grams of oxygen. What is the limiting reactant in the reaction between iron and oxygen to form iron oxide (FeO)? How many grams of FeO can be produced assuming complete reaction and no excess reactants?

Answer 1

To determine the limiting reactant and the amount of FeO produced, we first need to write the balanced chemical equation for the reaction:2 Fe + O2  2 FeONext, we need to find the molar mass of each reactant and product:Fe: 55.85 g/molO2: 32.00 g/molFeO: 71.85 g/mol  55.85 g/mol for Fe + 16.00 g/mol for O Now, we can calculate the moles of each reactant:moles of Fe = 20 g / 55.85 g/mol = 0.358 molmoles of O2 = 40 g / 32.00 g/mol = 1.25 molAccording to the balanced equation, 2 moles of Fe react with 1 mole of O2. We can now determine which reactant is limiting by comparing the mole ratios:Fe: 0.358 mol / 2 = 0.179O2: 1.25 mol / 1 = 1.25Since 0.179 is less than 1.25, Fe is the limiting reactant.Now, we can calculate the amount of FeO produced using the stoichiometry of the balanced equation:moles of FeO = 0.358 mol Fe *  2 mol FeO / 2 mol Fe  = 0.358 mol FeOFinally, we can convert the moles of FeO to grams:grams of FeO = 0.358 mol * 71.85 g/mol = 25.7 gTherefore, 25.7 grams of FeO can be produced assuming complete reaction and no excess reactants.