Question

For the following reaction, 33.7 grams of iron are allowed to react with 18.6 grams of oxygen gas . iron(s) + oxygen(g) yields to make iron(III) oxide(s) What is the maximum mass of iron(III) oxide that can be formed? What is the FORMULA for the limiting reagent? O2 What mass of the excess reagent remains after the reaction is complete?

Answer 1

number of mole = (given mass)/(molar mass)
molar mass of Fe = 55.8 g/mol
molar mass of O2 = 32 g/mol
number oof mole of O2 = 18.6/32
= 0.58 mole
number of mole of Fe = 33.7/55.8
= 0.6 mole

reaction taking place is
2Fe + 3O2 -- > 2Fe2O3
according to reaction
2 mole of Fe required 3 mole of O2
1 mole of Fe required 3/2 mole of O2
0.6 mole of Fe required (3/2)*0.6 mole of O2
0.6 mole of Fe required 0.9 mole of O2
but we have 0.58 mole of O2
so, O2 Is limiting reagent

3 mole of O2 give 2 mole Fe2O3
1 mole of O2 give 2/3 mole Fe2O3
0.58 mole of O2 give (2/3)*0.58 mole Fe2O3
number of mole of Fe2O3 formed = (2/3)*0.58
= 0.39 mole
mass of Fe2O3 formed = (number of mole)*(molar mass of Fe2O3)
molar mass of Fe2O3 = 159.7 g/mol
mass of Fe2O3 formed = 0.39*159.7
= 62.3 g

here, O2 is limiting reagent

3 mole of O2 react with 2 mole of Fe
0.58 mole of O2 react with (2/3)*0.58 mole of Fe
number of mole of Fe used = (2/3)*0.58
= 0.39 mole
number of mole of Fe are in excess = 0.6 - 0.39
= 0.21 mole
mass of Fe in excess = (number of mole of Fe are in excess)*(molar mass of Fe)
= 0.21*55.8
= 11.7 g