Excess and Limiting Reactants

All the reactions we've done up until now assume that all reactants are completely used up during reaction. This is not the case. Often, reactants are in excess; there is some left over.

Here is an example:

What mass of CS2 is produced when 18.0 g of C are reacted with 41.0 g of SO2?

5 C + 2 SO2-------> CS2 + 4 CO

What mass of excess reactant is left over?

Here try and find mass of CS2 by using the masses of the reactants C and SO2.

Mass of CS2 (based on C) 18.0 g C x 1 mole/ 12.0 x 1 CS2/5 moles C x 76.2 / 1 mole= 22.9 g CS2

Mass of CS2 (based on SO2)  41.0 g SO2 x 1 mole/ 64.1 x 1 CS2/2 SO2 x 76.2/ 1mole = 24.4 g CS2

Since there is too much SO2 and not enough C, SO2 is the excess reactant because in the reaction all the C will be used  up and the SO2 will be in excess.

In this reaction, C is the limiting reactant because it limits the amount of CS2 that can be formed, there is lots of SO2 and not enough C to make that much SO2.

Example #2. What mass of P4 is produced when 38.0 g of Ca3(PO4)2, 25.0 g of SiO2 and 9.70 g of C are reacted?

2 Ca3(PO4)2 + 6 SiO2 + 10 C -----> P4 + 6 CaSiO3 + 10 CO

Use the mass of each reactant and find the mass of P4.

Mass of P4( based on Ca3(PO4)2)38.0 g Ca3(PO4)2 x 1 mole/ 310.3 x 1 P4/2 Ca3(PO4)2 x 124.0/1 mole = 7.59 g P4

Mass of P4 ( based on SiO2) 25.0 g SiO2 x 1 mole/60.1 x 1 P4/ 6 SiO2 x 124.0/1mole = 8.60 g P4

Mass of P4 (based on C) 9.70 g C x 1mole/12.0 x 1 P4/ 10 C x 124/1mole= 10.0 g P4.

Ca3(PO4)2 produces the least amount of P4 = 7.59 g

The excess reactants are  SiO2 and C.

How many grams of excess reactant are left over?

38.0 g Ca3(PO4)2 x 1mole/ 310.3 x 6 SiO2/2 Ca3(PO4)2 x 60.1/1 mole = 22.1 g SiO2.

To calculate excess, subtract this number from the original amount.

25.0 - 22.1 = 2.9 g SiO2 in  excess.

38.0 g Ca3(PO4)2 x 1 mole/310.3 x 10 C/ 2 Ca3(PO4)2 x 12.0 / 1mole =7.35 g C

Subtract from original.

9.70 g- 7.35 = 2.35 g excess C







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