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Recap – Molar Mass and Moles Relative Atomic Mass – weighted average mass of all isotopes of a particular element relative to 12 C. Avogadro number – 6.022.

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Presentation on theme: "Recap – Molar Mass and Moles Relative Atomic Mass – weighted average mass of all isotopes of a particular element relative to 12 C. Avogadro number – 6.022."— Presentation transcript:

1 Recap – Molar Mass and Moles Relative Atomic Mass – weighted average mass of all isotopes of a particular element relative to 12 C. Avogadro number – x this is the number of atoms of 12 C in exactly g of 12 C. The mole – name we use when we have an Avogadro number of objects. 1 Amount in moles, n = mass / molar mass

2 2 Stoichiometry Q:What masses of Na and Cl 2 are needed to produce 100 g NaCl? 2Na + Cl 2  2NaCl Mass/g ? ?100 Molar mass Moles 1.71

3 3 Stoichiometry Q:What masses of Na and Cl 2 are needed to produce 100 g NaCl? 2Na + Cl 2  2NaCl Mass/g ? ?100 Molar mass Moles Use balanced equation

4 4 Stoichiometry Q:What masses of Na and Cl 2 are needed to produce 100 g NaCl? 2Na + Cl 2  2NaCl Mass/g Molar mass Moles m = n M

5 5 Limiting Reagent So far we have talked about reactions where we have exactly the right amounts of each reactant. Most of the time, however, we will not have enough of one reactant (reagent) to react with all of the other. The reagent which we are short of is called the limiting reagent – it determines how much product we get.

6 6 Limiting Reagent Q:5.00 g of H 2 and 10.0 g of F 2 are mixed. When the reaction ceases, what is the mass of each type of molecules present? H 2 +F 2  2HF Mass/g Molar mass Moles  F 2 is the limiting reagent – amount of F 2 will determine amount of product.

7 7 Limiting Reagent Q:5.00 g of H 2 and 10.0 g of F 2 are mixed. When the reaction ceases, what is the mass of each type of molecules present? H 2 +F 2  2HF Mass/g Molar mass Moles (initially) Moles (used) Moles (after react.) Mass (after react.)

8 8 % Yield So far we have assumed that all of the limiting reagent is used up – 100 % reaction – the reaction ‘has gone to completion’. This may not always be the case.

9 9 % Yield Q:In the reaction of 5.0 g of sodium with excess chlorine, the mass of sodium chloride obtained was 12.4 g. Calculate the % yield. 2Na+Cl 2  2NaCl Mass/g 5.0 Molar mass Mole

10 10 % Yield Q:In the reaction of 5.0 g of sodium with excess chlorine, the mass of sodium chloride obtained was 12.4 g. Calculate the % yield. 2Na+Cl 2  2NaCl Mass/g Molar mass Mole % yield = 12.4 / 12.7  100 = 97.6 %

11 By the end of this lecture, you should: −be able to perform stoichiometric calculations including: −determining the mass of product given the mass of starting materials −determine the mass of starting materials needed to give a required mass of product −identifying the limiting reagent and using this in the calculations −determine the % yield in a reaction given the mass of starting materials −be able to complete the worksheet (if you haven’t already done so…) 11 Learning Outcomes:

12 12 Questions to complete for next lecture: 1.Write a balanced equation for the complete combustion of carbon to form carbon dioxide. 2.How many moles of carbon are there in 100 g of carbon? 3.If the combustion is carries out in air, which is the limiting reagent? 4.How many moles of carbon dioxide would this produce on complete combustion of 100 g of carbon? 5.What mass of carbon dioxide is produced from the complete combustion of 100 g of carbon? 6.If 348 g of carbon dioxide is isolated from complete combustion of 100 g of carbon, what % yield does this represent?


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