Stoichiometry

Stoichiometry

Adjusting To Reality

Adjusting To Reality

This is not the entire story. In reality, you never have the exact amounts of both reactants you need. At the end of the reaction, one reactant has been completely consumed and there is some “left over” of the other reactant.

Let’s take a second look at the reaction between PbS and O2.

2PbS + 3O2 2PbO + 2SO2

Since oxygen is not costly (free in our atmosphere), it’s usually the reactant there is plenty of and only a certain amount of lead II sulfide would have been purchased for this reaction.

What would our BCA table look like if we had 0.40 moles of lead (II) sulfide reacting with an abundance (excess) of oxygen?

Adjusting To Reality Equation: 2PbS + 3O2 2PbO + 2SO2

Before: .40 mol xs mol 0 mol 0 mol

Change - .40 mol - xs mol +.40 mol +.40 mol ____________________________________After 0 mol xs mol .40 mol .40 mol

Excess is written (xs) and indicates there is some reactant remaining. Here, PbS is completely consumed and some O2 remains after the reaction is complete.

Further RealityWhat if only a certain amount of each reactant were available?

25.50 g of oxygen reacts with 114.85 g lead (II) sulfide producing lead (II) oxide and sulfur dioxide. What mass of lead (II) oxide would be produced?

We cannot directly measure moles, so the reactant amounts are given in grams. In order to use our BCA table (for mole ratios) we need the amounts in moles. Using molar mass, we convert the mass of the reactants to moles of reactants.

Mass To Moles, “Molar Mass”

25.50 g O2 x 1 mol O2_ = .80 mol O2

32.0 g O2

114.85 g PbS x _1 mol PbS_ = .48 mol PbS 239.27g

PbS

A Second Look

Equation: 2PbS + 3O2 2PbO + 2SO2

Before: .48 mol .80 mol 0 mol 0 mol

Change -__ mol - __mol +__mol+__mol _______________________________

After __mol __mol __mol __ mol

Limiting & Excess Reactants

Our first task is to find out which reactant will be completely consumed (limiting reactant) and which reactant will have some remaining after the reaction is complete (reactant in excess). We will use mole ratios from the BCA table for this task.

Limiting & Excess Reactants

.80 mol O2 x 2 mol PbS = .53 mol PbS 3 mol O2

We need .53 mol of PbS to completely react .80 mol of O2

.48 mol PbS x 3 mol O2 = .72 mol O2

2 mol PbSWe need .72 mol of O2 to completely react .48 mol of PbS

Evaluating Our Answers

We need 0.53 mol PbS to completely burn 0.80 mol O2. We only have 0.48 mol

PbS. Not all of the O2 will be

“consumed”. The reaction will stop when the PbS has run out. This tells us the PbS will limit the reaction (limiting reactant) and some oxygen will remain after the reaction is complete (reactant in excess).

Amount of Reactant Remaining

We need 0.72 mol O2 to completely

react with 0.48 mol PbS. We have

0.80 mol O2. 0.08 mol O2 will remain

after all of the PbS has been

consumed.

Importance of Limiting Reactant

The PbS limits the reaction. PbS

“runs out” before all the O2 is

consumed. PbS is the reactant that determines how much product will be produced.

Putting It All Together!

Equation: 2PbS + 3O2 2PbO + 2SO2

Before: .48 mol .80 mol 0 mol 0 mol

Change -.48 mol - .72 mol + .48 mol + .48 mol ___________________________________After: 0 mol .08 mol .48 mol .48 mol