Question

What mass of iron is produced from adding 80.0 g of iron(II) oxide (71.85 g/mol) to 20.0 g of magnesium metal? FeO (l) + Mg (l) ----------> Fe (l) + MgO (s) Identify the limiting reactant, mass of iron produced and the mass in grams of the excess reactant used in this reaction. Remember to report the correct number of significant figures and units where appropriate. You need to show work in order to receive credit. Zero points will be awarded if logical work/explanation is not given. You must show your work by typing it into the provided test box. I will not look to your scratch paper for work.

Answer 1

46.0g of Iron are produced

Step-by-step explanation:

Based on the chemical reaction:

FeO(l) + Mg(l) → Fe(l) + MgO(s)

1 mole of Iron (II) oxide reacts per mole of Mg to produce 1 mole of iron

To solve this question we need to convert each mass of reactant to moles using its respectives molar masses in order to find limitng reactant. Moles of limiting reactant = Moles of iron produced:

Moles FeO (Molar mass: 71.85g/mol):

80.0g * (1mol / 71.85g) = 1.11moles FeO

Moles Mg (Molar mass: 24.305g/mol)

20.0g * (1mol / 24.305g) = 0.823 moles Mg

As moles of Mg < Moles FeO, Mg is limiting reactant and the moles of Fe are 0.823 moles.

The mass of Iron produced is:

0.823 moles Fe * (55.845g/mol) =

46.0g of Iron are produced