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Stoichiometry, Limiting Reactants and Percent Yield

Stoichiometry, Limiting Reactants and Percent Yield. How many grams of carbon dioxide will be produced when 11.6 g of butane (C 4 H 10 ) burns with oxygen?. 11.6 g ? g. 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O.

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Stoichiometry, Limiting Reactants and Percent Yield

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  1. Stoichiometry,Limiting Reactants and Percent Yield

  2. How many grams of carbon dioxide will be produced when 11.6 g of butane (C4H10) burns with oxygen? 11.6 g ? g 2 C4H10 + 13 O2 8 CO2 + 10 H2O 2 mole 8 mole 11.6 g C4H10 1 mole C4H10 X -------------------- 58.1 g C4H10 8 mole CO2 X---------------- 2 mole C4H10 44.0 g CO2 X --------------- = 1 mole CO2 35.1 g CO2

  3. Excess/Limiting Reactants • The limiting reactant is the reactant present in the smallest stoichiometric amount • In other words, it’s the reactant you’ll run out of first! • The excess reactant will be left over after the reaction.

  4. Limiting Reactants Example a) 11.6 grams of magnesium react with 54.6 grams of nitric acid. Hydrogen and magnesium nitrate are produced. How many grams of hydrogen are produced? 11.6 g 54.6 g x g Mg + 2 HNO3 ----> H2 + Mg(NO3)2

  5. Method 1 Here are the 2 possible calculations (you really only need to do one of these): 1 mol Mg 2 mol HNO3 63.0 g HNO3 11.6 g Mg X ----------- X --------------- X -------------- = 60.1 g HNO3 24.3 g Mg 1 mole Mg 1 mole HNO3 We can’t use all the Mg – there’s not enough HNO3 to react OR 1 mole HNO3 1 mole Mg 24.3 g Mg 54.6 g HNO3 X-------------- X -------------- X ----------- = 10.5 g Mg 63.0 g HNO3 mole HNO3 mole Mg If we use all the HNO3 we’ll still have Mg left over

  6. Method 1 cont. ALWAYS USE THE LIMITING REACTANT TO FIND THE REQUIRED! 1 mole HNO3 1 mole H2 2.02 g H2 54.6 g HNO3 X --------------- X --------------- X ---------- = .875 g H2 63.0 g HNO3 2 mole HNO3 1 mole H2

  7. Method 2 AND The smaller answer is correct – 0.875g of H2 is formed. HNO3 is the limiting reactant and Mg is the excess reactant We know this because we used HNO3 to find the correct answer

  8. Part b… • How much magnesium is left over after the reaction? (you may have already done the first step if you happened to do this calculation to find the limiting reactant in method 1) 1 mole HNO3 1 mole Mg 24.3 g Mg 54.6 g HNO3 X-------------- X -------------- X ----------- = 10.5 g Mg 63.0 g HNO3 mole HNO3 mole Mg 11.6g Mg – 10.5 g Mg = 1.1 g Mg is left over!

  9. Theoretical Yield • The theoretical yield is the amount of product that can be made • This is what we calculate • The actual yield is the amount one actually produces and measures • Either measured in the lab or given in the problem

  10. Actual Yield Theoretical Yield Percent Yield = x 100% Percent Yield A comparison of the amount actually obtained to the amount it was possible to make

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