Determining Limiting Reactants and Products produced. [ENDORSED]

ERIKTORRESDisc3C
Posts: 18
Joined: Wed Sep 21, 2016 2:58 pm

Determining Limiting Reactants and Products produced.

I had little experience with Limiting Reactant Calculations in High School since a lot of this material was not covered in my High School classes. I do not know how to set up and solve for products from limiting reactants and would like a simplified explanation of the steps to go about the mathematical process of solving chemistry word problems.
"Boron Oxide Reacts with Magnesium to form solid Boron and solid Magnesium Oxide. If 125 Kg of Boron Oxide reacts with 125 Kg of Magnesium Oxide, how much solid Boron will result from the reaction of Boron Oxide with Magnesium?"

Sarah_Heesacker_3B
Posts: 21
Joined: Wed Sep 21, 2016 2:57 pm

Re: Determining Limiting Reactants and Products produced.  [ENDORSED]

If you have not already, watching the module on Limiting Reactant Calculations is extremely helpful and gives these steps to solving these types of problems:
1. Identify reactants and products (In your example, the reactants are Boron Oxide and Magnesium. The products are Boron and Magnesium Oxide.)
2. Write and balance equation for reaction (I am not sure what kind of Boron Oxide the problem refers to, but the internet says Boron trioxide is the most common. $B_{2}O_{3}(s)+3Mg(s)\rightarrow 2B(s)+3MgO(s)$)
3. Calculate molar mass of each reactant and product ($B_{2}O_{3}$
=69.62g/mol, Mg=24.31g/mol, B=10.81g/mol, MgO=40.31)
4. Convert known masses of reactants and products to moles (In this case, convert kilograms to grams, then divide my molar mass. $1.795\times 10^{3}$ mol $B_{2}O_{3}$ and $5.142\times 10^{3}$ mol Mg)
5. Compare calculated moles from Step 4 to required moles in Step 2 to determine if there is a limiting reactant. ($1.795\times 10^{3}$ mol $B_{2}O_{3}$ requires 3 times that amount of Mg since the molar ratio is 1:3. Therefore you would need $5.385\times 10^{3}$ mol of Mg but you only have $5.142\times 10^{3}$ mol Mg, so Mg is the limiting reactant)
6. Based on moles of the limiting reactant, calculate the moles of the product than can form. ($5.142\times 10^{3}$ mol Mg] multiplied by 2/3 is $3.428\times 10^{3}$ mol B since the molar ratio is 2:3.)
7. Convert moles of product to grams using molar mass. This is theoretical yield. ($3.428\times 10^{3}$ mol B multiplied by the molar mass of B is $3.71\times 10^{4}$g which is your final answer!)

I hope this helps!

ERIKTORRESDisc3C
Posts: 18
Joined: Wed Sep 21, 2016 2:58 pm

Re: Determining Limiting Reactants and Products produced.

Thank You SO MUCH!