4Al + 3O2 —> 2Al2 O3 10.7g of powdered Al is placed into a container containing 10.7g O2. What is the limiting reactant? How many grams of aluminum oxide can be produced? Calculate the mass of excess reactant that remains after the reaction is complete

Limiting reactant : Al

Al2O3 produced = 18.36 g

mass O2 remains = 11.904 g

Given

Reaction

4Al + 3O2 —> 2Al2O3

10.7 g O2

Required

Limiting reactant

mass Al2O3

Mass of excess reactant remains

Solution

mol Al : 10.7 : 27 = 0.396

mol O2 = 10.7 : 32 = 0.334

1. limiting reactant

mol : coefficient of Al : O2 = 0.396/4 : 0.334/3 = 0.099 : 0.111

Al as a limitng reactant(smaller ratio)

2. mass Al2O3

mol Al2O3 based on Al : 2/4 x 0.396 = 0.18

mass Al2O3(MW=102 g/mol) = 0.18 x 102 = 18.36 g

3. mass O2 remains

mol O2 reacted : 3/4 x 0.396 = 0.297

mol O2 remains = 0.669 - 0.297 =0.372

mass O2 remains = 0.372 x 32 = 11.904 g

h let me think

the answer is symmetrical : )