Question

Part IV. Limiting Reactants! A Challenge Problem!

Answer 1

Answer:

a. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

b. Fe2O3 is the limiting reactant.

c. 6.30 grams Fe

d. 52.6 %

Explanation:

Step 1: Data given

Mass of iron(III) oxide Fe2O3 = 9.00 grams

Mass of aluminium = 4.00 grams

Molar mass Fe2O3 = 159.69 g/mol

Aluminium molar mass = 26.98 g/mol

Step 2: The balanced equation

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

Step 3; Calculate Moles

Moles = mass / molar mass

Moles Fe2O3 = 9.00 grams / 159.69 g/mol

Moles Fe2O3 = 0.0564 moles

Moles Al = 4.00 grams / 26.98 g/mol

Moles Al = 0.148 moles

Step 4: Calculate limiting reactant

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

Fe2O3 is the limiting reactant. It will completely be consumed (0.0564 moles).  Al is in excess. There will react 0.0564*2 = 0.1128 moles

There will remain 0.148 - 0.1128 = 0.0352 moles Al

Step 5: Calculate moles Fe

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

For 0.0564 moles Fe2O3 we'll have 2*0.0564 = 0.1128 moles Fe

Step 6: Mass of Fe

Mass Fe = 0.1128 moles * 55.845 g/mol

Mass Fe = 6.30 grams

Step 7: If you carried out this reaction and it actually produced 0.475 mL of molten iron (r = 6.98 g/mL), what is the percent yield of this reaction?

Density = mass / volume

Mass = density * volume

Mass = 6.98 g/mL * 0.475 mL

Mass = 3.3155 grams

Percent yield = (actual mass / theoretical mass) * 100%

Percent yield = (3.3155 /6.30 ) * 100 %

Percent yield = 52.6 %