In any type of chemical reaction, the amount of product that can be produced is determined by the reactant which is in the smallest amount.
In any type of chemical reaction, the amount of product that can be produced is determined by the reactant which is in the smallest amount. In other words once the reactant that is present in the smallest amount is completely consumed the reaction will stop.
Definition Limiting Reactant = a chemical which limits the amount of product that can be obtained in the combining process.
Definition Limiting Reactant = a chemical which limits the amount of product that can be obtained in the combining process. Another way to put it is that the Limiting Reactant is the substance that is used up FIRST in a chemical reaction.
But when determining which element is the limiting reactant or how much product can be produced - the Stoichiometry is the SAME!
But when determining which element is the limiting reactant or how much product can be produced - the Stoichiometry is the SAME! The steps will still be: 1. Balance the equation. 2. Convert grams to moles. 3. Convert moles to moles. 4. Convert moles to grams.
But when determining which element is the limiting reactant or how much product can be produced - the Stoichiometry is the SAME! The steps will still be: 1. Balance the equation. 2. Convert grams to moles. 3. Convert moles to moles. 4. Convert moles to grams. You just have to do it TWICE!
Let s take for example: Calculate the mass, in grams, of NaCl possible if 45 g of Na reacts with 58 g of Cl2. Na + Cl2 NaCl
Let s take for example: Calculate the mass, in grams, of NaCl possible if 45 g of Na reacts with 58 g of Cl2. Na + Cl2 2 Na + Cl2 2 NaCl NaCl
Let s take for example: Calculate the mass, in grams, of NaCl possible if 45 g of Na reacts with 58 g of Cl2. Na + Cl2 2 Na + Cl2 2 NaCl 45 g Na 1 mol Na 22.99 g Na 2 NaCl 2 Na NaCl 58.44 g NaCl = 114 g 1 mol NaCl NaCl
Let s take for example: Calculate the mass, in grams, of NaCl possible if 45 g of Na reacts with 58 g of Cl2. Na + Cl2 2 Na + Cl2 2 NaCl 45 g Na 58 g Cl2 1 mol Na 22.99 g Na 2 NaCl 2 Na 1 mol Cl2 2 NaCl 70.90 g 1 mol Cl2 Cl2 NaCl 58.44 g NaCl = 114 g 1 mol NaCl NaCl 58.44 g NaCl = 95 g 1 mol NaCl NaCl
Let s take for example: Calculate the mass, in grams, of NaCl possible if 45 g of Na reacts with 58 g of Cl2. Na + Cl2 2 Na + Cl2 2 NaCl 45 g Na 1 mol Na 22.99 g Na 2 NaCl 2 Na NaCl 58.44 g NaCl = 114 g 1 mol NaCl NaCl 58 g Cl2 1 mol Cl2 2 NaCl 58.44 g NaCl = 95 g 70.90 g 1 mol 1 mol NaCl NaCl Cl2 Cl2 So the greatest amount of NaCl that can be produced is 95 g,
Let s take for example: Calculate the mass, in grams, of NaCl possible if 45 g of Na reacts with 58 g of Cl2. Na + Cl2 2 Na + Cl2 2 NaCl 45 g Na 1 mol Na 22.99 g Na 2 NaCl 2 Na NaCl 58.44 g NaCl = 114 g 1 mol NaCl NaCl 58 g Cl2 1 mol Cl2 2 NaCl 58.44 g NaCl = 95 g 70.90 g 1 mol 1 mol NaCl NaCl Cl2 Cl2 So the greatest amount of NaCl that can be produced is 95 g, because we run out of Cl2.
That really wasn t that bad.
Let s try again: What is the maximum mass, in grams, of H3PO4 possible if 100 g of P4O10 reacts with 200 g of H2O? P4O10 + H2O H3PO4
Let s try again: What is the maximum mass, in grams, of H3PO4 possible if 100 g of P4O10 reacts with 200 g of H2O? P4O10 + 6 H2O P4O10 + H2O 4 H3PO4 H3PO4
Let s try again: What is the maximum mass, in grams, of H3PO4 possible if 100 g of P4O10 reacts with 200 g of H2O? P4O10 + 6 H2O P4O10 + H2O 4 H3PO4 H3PO4 100 g P4O10 1 mol P4O10 283.88 g P4O10 4 mol H3PO4 1 mol P4O10 = 1.41 mol H3PO4
Let s try again: What is the maximum mass, in grams, of H3PO4 possible if 100 g of P4O10 reacts with 200 g of H2O? P4O10 + 6 H2O P4O10 + H2O 4 H3PO4 H3PO4 But let s see a secret! 100 g P4O10 1 mol P4O10 283.88 g P4O10 4 mol H3PO4 1 mol P4O10 = 1.41 mol H3PO4 200 g H2O 1 mol H2O 4 mol H3PO4 18.02 g H2O 6 mol H2O = 7.40 mol H3PO4
Let s try again: What is the maximum mass, in grams, of H3PO4 possible if 100 g of P4O10 reacts with 200 g of H2O? P4O10 + 6 H2O P4O10 + H2O 4 H3PO4 H3PO4 But let s see a secret! 100 g P4O10 1 mol P4O10 283.88 g P4O10 4 mol H3PO4 1 mol P4O10 = 1.41 mol H3PO4 200 g H2O 1 mol H2O 4 mol H3PO4 18.02 g H2O 6 mol H2O = 7.40 mol H3PO4 So the most H3PO4 that could be made is 1.41 moles.
Let s try again: What is the maximum mass, in grams, of H3PO4 possible if 100 g of P4O10 reacts with 200 g of H2O? P4O10 + 6 H2O P4O10 + H2O 4 H3PO4 H3PO4 But let s see a secret! 100 g P4O10 1 mol P4O10 283.88 g P4O10 4 mol H3PO4 1 mol P4O10 = 1.41 mol H3PO4 200 g H2O 1 mol H2O 4 mol H3PO4 18.02 g H2O 6 mol H2O = 7.40 mol H3PO4 So the most H3PO4 that could be made is 1.41 moles. So 1.41 mol 98.00 g/mol = 138 g possible.
That maximum amount of product that can be produced, if all of the limiting reactant is consumed, is called the THEORETICAL YIELD.
That maximum amount of product that can be produced, if all of the limiting reactant is consumed, is called the THEORETICAL YIELD. But sometimes the theoretical yield is not achieved. Sometimes all of the limiting reactant is not completely consumed because side reactions or even reverse reactions may occur.
So to that end, chemists calculate what is called the PERCENT YIELD.
So to that end, chemists calculate what is called the PERCENT YIELD. Percent Yield Actual Yield Theoretical Yield x 100 where the Actual Yield is determined in the lab by measuring the amount of product produced.
Take that last example - we determined that we could only make 138 grams of H3PO4.
Take that last example - we determined that we could only make 138 grams of H3PO4. But what if only 126 grams was produced. What would be the percent yield?
Take that last example - we determined that we could only make 138 grams of H3PO4. But what if only 126 grams was produced. What would be the percent yield? Percent Yield 126 g 138 g x 100 91.3 %
Now it is YOUR TURN: What is the maximum mass, in grams, of Ag possible if 100 g of Cu reacts with 100 g of AgNO3? AgNO3 + Cu Cu(NO3)2 + Ag And what would be the percent yield if only 59 grams of Ag are produced?