Limiting Reagents & Moles
In a Chemistry lab you are asked to figure out how many moles of Al2O3 can be produced from the reaction of 10.0g of Al and 19.0g of O3.
How many moles of Al2O3 can be produced from 10.0g Al and 19.0g O3?
1. Read thoroughly the problem.
2. Write out a balanced chemical reaction/equation.
4 Al + 3 O3 ——-> 2 Al2O3 LR ration = 4:2
3. Think about each quantity of reactants compared to the products. The amount of product is determined by the quantity of the Limiting Reactant no matter how much of the ingredients/chemicals you have.
4. Identify the Limiting Reactants: Aluminum (10.0g Al) How do you know?
10.0gAl (1mol Al/26.9g Al) = 0.371mol Al
19.0g O3 (1mol O3/16gx2 O3) = 0.59 4mol O2
The quantity of moles of Al is less than O2. Therefore, it is the limited reactant.
5. Use the LR of (0.371mol Al) to find the moles of Al2O3.
(0.371mol Al) x (2mol Al2O3/ 4mol Al) = 0.1853 mol of Al2O3
6. Check your answer.