Stoiciometry & Limiting reagent

Question

In the reaction A + 2B ---> 3C + D , If 93g of A and 85g of B are combined, which reactant is the limiting reagent, and what is the theoretical yield of C in grams? (MM(A)= 55g/mole, MM(B)= 48g/mole, MM(C)= 37g/mol)

Answer 1

So you need to approach this in two different ways: calculating C from A and calculating C from B.

 

Calculating C from A:

93gA* 1molA/55g* 3 mol C/ 1 mol A * 37g/mol C = 187.7 g C

 

Calculating C from B:

85g B* 1mol B/48g* 3 mol C/ 2 mol B * 37g/mol C = 98.3 g C

 

Since there is less C calculated from B, B is the limiting reagent. 98.3 g C is the theoretical yield

Answer 2

Hi Catherine!

 

First, let's convert the mass of A and B into moles

A: (93g)/(55g/mol) = 1.69 mol

B: (85g)/(48g/mol) = 1.77 mol

Next, we can use the stoichiometric coefficients to determine the limiting reagent

1mol of A produces 3mols of C (1:3 ratio of coefficients)

2 mol of B produces 3 mol of C (2:3 ratio of coefficients)

 

We can then solve the number of moles of C that can be produced by each reagent and see which is less - This will be the limiting reagent.

For A: 1.69mol * (3mol C/1mol A) = 5.07mol C

For B: 1.77mol * (3mol C/2mol B) = 2.655 mol C

 

Because the initial amount of B produces less of C than A can, we know B is the limiting reagent!

To find the theoretical yield of C in grams, take the lower amount that can be produced by the initial reagents (2.655mol from B) and multiply it by the MM to find the mass

2.655mol*(37g/mol) = 98.235g

Theoretical Yield = 98g

 

Let me know if you have any questions!