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 %
). Hope this helps.
