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Limiting Reagents and Percent Yield

Limiting Reagents and Percent Yield. Using Stoichiometry. Chemistry Joke. Q: Why were copper and titanium the cheerleader’s favorite elements?. A: Because she knew she was a Cu Ti!. Limiting Reagent— t he reactant that controls the quantity of product formed.

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Limiting Reagents and Percent Yield

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  1. Limiting Reagents and Percent Yield Using Stoichiometry

  2. Chemistry Joke Q: Why were copper and titanium the cheerleader’s favorite elements? A: Because she knew she was a Cu Ti!

  3. Limiting Reagent—the reactant that controls the quantity of product formed (We use the terms reagent and reactant interchangeably.) A real world example…

  4. You want to make a lot of chocolate chip cookies! You have – 4 lbs of butter, 2 lbs of salt, 1 gallon of vanilla extract, 8 lbs of chocolate chips, 20 lbs flour, 15 lbs of sugar, 10 lbs of baking soda, and TWO eggs. • What is the limiting reagent???

  5. Limiting Reactant – the reactant that is used up completely. • Excess Reactant – the reactant that is not used up completely in a reaction. There’s some left over.

  6. Finding the Limiting Reagent • All limiting reagent problems will have more than one reactant amount. • If there is more than one given amount of reactant, you MUST find the limiting reagent.

  7. Step 1 • If the given amount of each reactant is not in moles-then convert each value into moles • Step 2 • Divide each number of moles by the coefficient for that reactant in the balanced equation.

  8. Step 3 The smaller amount of reactant after dividing is the limiting reactant .

  9. An Example… • Zn + 2HCl  ZnCl2 + H2 • What is the limiting reactant when 12.1 g Zn reacts with 2.65 g HCl? Notice the two “given” reactants.

  10. Change into moles. • Divide by coefficient. • Zn + 2HCl  ZnCl2 + H2 12.1 g Zn 1 mol Zn = 0.185 mol Zn 1 mol Zn = 0.185 65.38 g Zn = 0.0727 molHCl 2 molHCl 2.65 g HCl 1 molHCl = 0.0363 36.46 g HCl 0.0363 < 0.185 therefore, HCl is limiting.

  11. Another Example… • Suppose 6.70 g Na reacts with 3.20 g Cl2. Which is the limiting reactant? 2 Na + Cl2 2 NaCl • Divide by coefficient. Change into moles.

  12. Finding the Amount of Product • First, find the limiting reagent. • Then, use stoichiometry starting with the MASS of the LIMITING REAGENT!

  13. For Example… • A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the reaction: Mg(OH)2+ 2 HCl→MgCl2 + 2 H2O How much MgCl2 will be produced? • There are 2 given reactants so I must first find the limiting one. • I will then use the mass of the limiting reactant to find how much product will be produced.

  14. For Example… • A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the reaction: Mg(OH)2+ 2 HCl→MgCl2 + 2 H2O How much MgCl2 will be produced?

  15. Percent Yield • Theoretical Yield – the maximum, calculated amount of product that can be produced • This number comes from stoichiometry!! • Actual Yield – the measured amount of product (experimental yield) • This number comes from lab experiments. • Percent Yield = actual yield theoretical yield x 100

  16. An Example… • When 84.8 g of iron (III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the % yield? 54.3 g 84.8 g Fe2O3 + 3 CO → 2 Fe + 3CO2 • Label the problem: Notice that one number in the problem is the amount of a reactant, and the other number in the problem is the actual yield of a product.

  17. Use stoichiometry to find the theoretical yield in grams. • Start with the amount of reactant. The compound that was actually produced is your unknown. 54.3 g actual yield 84.8 g Fe2O3 + 3 CO → 2 Fe + 3CO2 • Then, divide the actual yield by your answer and multiply by 100 to find percent yield.

  18. Chemistry Joke It’s weird that uranium and iodide don’t combine more in nature… Because I think U and I would be great together!!

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