What masses of iron(iii) oxide and aluminum must be used to produce 15.0 g iron? what is the maximum mass of aluminum oxide that could be produced?

1) Chemical reaction:

Fe2O3 + 2Al ---> Al2O3 + 2Fe

2) molar ratios

1 mol Fe2O3 : 2Al : 1 mol Al2O3 : 2 mol Fe

3) Convert 15.0 g of iron into moles

atomic mass Fe = 55.8 g/mol

moles = mass in grams / atomic mass = 15.0 g / 55.8 g/mol = 0.269 mol

4) Use proportions to determine the moles of Fe2O3, Al, and Al2O3

a) 1mol Fe2O3 / 2 mol Fe = x / 0.269 mol Fe

x =

=> x = 0.269 mol Fe * 1 mol Fe2O3 / 2 mol Fe = 0.134 mol Fe2O3

b) 2 mol Al / 2 mol Fe = x / 0.269 mol Fe

=> x = 0.269 mol Al

c) 2 mol Fe / 1 mol Al2O3 = 0.269 mol Fe / x

=> x = 0.269 mol Fe * 1 mol Al2O3 / 2 mol Fe

x = 0.134 mol Al2O3

5) Convert moles to grams

a) Fe2O3

molar mass Fe2O3 = 2* 55.8 g/mol + 3*16g/mol = 159.6 g/mol

mass = molar mass * number of moles

mass = 159.6 g/mol * 0.134 mol = 21.4 g

b) Al

atomic mass = 27.0 g/mol

mass = number of moles * atomic mass = 0.269 mol * 27.0 g/mol = 7.26 g

c) Al2O3

molar mass = 2 * 27.0 g/mol + 3*16.0 g/mol = 102.0g/mol

mass Al2O3 = numer of moles * molar mass = 0.134 mol * 102.0 g/mol = 13.7 g

Answers:

21.4 g Fe2O3

7.26 g Al

13.7 g Al2O3