# Stoichiometry Adjusting To Reality Adjusting To Reality  This is not the entire story. In reality, you never have the exact amounts of both reactants.

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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 O 2.

2PbS + 3O 2  2PbO + 2SO 2  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 + 3O 2  2PbO + 2SO 2 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 O 2 remains after the reaction is complete.

Further Reality  What 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 O 2 x 1 mol O 2 _ =.80 mol O 2 32.0 g O 2  114.85 g PbS x _1 mol PbS_ =.48 mol PbS 239.27g PbS

A Second Look  Equation: 2PbS + 3O 2  2PbO + 2SO 2 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 O 2 x 2 mol PbS =.53 mol PbS 3 mol O 2 We need.53 mol of PbS to completely react.80 mol of O 2 .48 mol PbS x 3 mol O 2 =.72 mol O 2 2 mol PbS We need.72 mol of O 2 to completely react.48 mol of PbS

Evaluating Our Answers WWe need 0.53 mol PbS to completely burn 0.80 mol O 2. We only have 0.48 mol PbS. Not all of the O 2 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 WWe need 0.72 mol O 2 to completely react with 0.48 mol PbS. We have 0.80 mol O 2. 0.08 mol O 2 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 O 2 is consumed. PbS is the reactant that determines how much product will be produced.

Putting It All Together!  Equation: 2PbS + 3O 2  2PbO + 2SO 2 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

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