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**Section 12.3 - “Limiting” Reagent**

The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make The limiting reactant determines how much of the excess reactant is used

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**How do you find out which is limited?**

Do two stoichiometry problems. The one that makes the least amount of product is the limiting reagent.

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**If 10. 6 g of copper reacts with 3**

If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of product (copper (I) sulfide) will be formed? 2Cu + S ® Cu2S Cu is Limiting Reagent 1 mol Cu 1 mol Cu2S g Cu2S 10.6 g Cu 63.55g Cu 2 mol Cu 1 mol Cu2S = 13.3 g Cu2S = 13.3 g Cu2S 1 mol S 1 mol Cu2S g Cu2S 3.83 g S 32.06g S 1 mol S 1 mol Cu2S = 19.0 g Cu2S

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**How much excess reagent will remain?**

Another example: If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper (grams) will be produced? How much excess reagent will remain? Hint: Use the REACTANT you determined to be the LR and solve for grams of the other reactant Then subtract this answer from the grams of excess reactant given in the problem

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**What is Yield? The amount of product made in a chemical reaction.**

There are three types: 1. Actual yield- what you get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield = Actual Theoretical x 100

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Example: 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu What is the actual yield? (from the problem) What is the theoretical yield? (Use grams of Al to determine grams of Cu) What is the percent yield?

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**Percent yield tells us how “efficient” a reaction is. **

Details on Yield: Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring

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Example Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 9.9g Pb(NO3)2 is heated to give 5.5g PbO 2Pb(NO3)2(s) 2PbO(s) + 4NO2(g) + O2(g) In a decomposition reaction, you don’t have a limiting reactant.

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Chapter 8 “Stoichiometry” Mr. Mole. Section 8.2 The Arithmetic of Equations u OBJECTIVES: Interpret balanced chemical equations in terms of: a) moles,

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