what mass of iron(III) oxide must be used to produce 10.6g iron?

The mass of iron (II) oxide that must be used to produce 10.6 g of iron would be 14.37 grams.Stoichiometric problemIron is produced from iron (II) oxide by reacting with carbon monoxide according to the following equation:[tex]Fe_2O_3 + 3CO -- > 2Fe + 3CO_2[/tex]The equation already obeys the law of conservation of atoms. Hence. it is a balanced equation.The mole ratio of iron (III) oxide that reacts and the iron that is produced according to the equation is 1:2.Now, in a particular reaction involving iron (III) oxide and carbon monoxide, 10.6 grams of iron were produced. The amount of iron (III) oxide used can be obtained using the mole ratio from the balanced equation of the reaction.Recall that: mole = mass/molar massMolar weight of iron = 56 g/molMole of 10.6 g of iron = 10.6/56                                     = 0.19 molesSince the mole ratio is 1:2, the equivalent mole of iron (III) oxide that reacts can be obtained:                   0.19/2 = 0.09 molesMolar mass of iron (III) oxide = 159.69 g/molMass of 0.09 mol iron (III) oxide = 0.09 x 159.69                                                     = 14.37 gramsThus, the amount of iron (III) oxide that must be used to produce 10.6 g of iron would be 14.37 grams.More on stoichiometric problems can be found here: https://brainly.com/question/28297916#SPJ1