1.  If you have 3.2 g of magnesium to start with (and plenty of oxygen) how many grams of MgO will be produced in this reaction? Mg + O 2  MgO

2. Unit 8 - Stoichiometry LIMITING REACTANT

3. Back to our Cake Analogy  1 cake mix + 3 eggs + 1 cup water  1 cake  If you have 2 mixes, 3 eggs and 5 cups of water, how many cakes can you make?  In this case we would call the eggs the limiting reactant. We can only make one cake because we don’t have enough eggs to match the other materials that are available. In this case we would make 1 cake and have 1 mix and 4 cups water left over.

4. Limiting reactant in a chemical Rxn 2 H 2 + O 2  2 H 2 O If you have 2 moles H 2 and 3 moles O 2, which is the limiting reactant? LR- H 2 You only need one mole of O 2 to react with the 2 moles of H 2 (you have an excess of 2 mol O 2 [3-1=2]). ER- O 2  2 mol H 2 X 1 mol O 2 = 1 mol O 2 2 mol H 2

5. Calculating Limiting reactant Always in moles!  7.6 mol H 2 X 1 mol O 2 = 3.8 mol O 2 2 mol H 2 The 3.5 mol O 2 you have is smaller than the 3.8 mol O 2 that you would need to react with all the H 2 so the O 2 is the limiting reactant. 2 H 2 + O 2  2 H 2 O If you have 7.6 moles H 2 and 3.5 moles O 2, which is the limiting reactant? Needed to react with 7.6 mol H 2

6.  Remember to balance the equation first!

7. Calculating Limiting reactant Always in moles!  2.0 mol HF X 1 mol SiO 2 = 0.50 mol SiO 2 4 mol HF The 4.5 moles SiO 2 you are given are MORE than enough to react with the 2.0 mol HF so HF is our limiting reactant SiO 2 + HF  SiF 4 + H 2 O If you have 2.0 mol HF and 4.5 mol SiO 2, which is the limiting reactant? 4 2

8. Calculating Limiting reactant Always goes through moles! 15 g HCl X 1 mol HCl = 0.41 mol HCl 36.5g HCl HCl + Ca(OH) 2  CaCl 2 + H 2 O If you have 15 g HCl and 12 g Ca(OH) 2, which is the limiting reactant? 2 12 g Ca(OH) 2 X 1 mol Ca(OH) 2 = 0.16 mol Ca(OH) 2 74 g Ca(OH) 2 X 1 mol Ca(OH) 2 = 0.205 mol Ca(OH) 2 2 mol HCl The Ca(OH) 2 is the limiting reactant because we do not have enough to react with all of the HCl we are given.

9. Using limiting reactant for stoichiometry  Zn + HCl  ZnCl 2 + H 2 If you are given 70 g Zn and 71 g HCl, what will be the mass of the salt produced? 2 71 g HCl X 1 mol HCl = 1.95 mol HCl 36.5g HCl 70 g Zn X 1 mol Zn = 1.07 mol Zn 65.4 g Zn X 2 mol HCl = 2.14 mol HCl 1 mol Zn The HCl is the limiting reactant because we do not have enough HCl to react with all of the Zn we are given.

10. Using limiting reactant  Zn + HCl  ZnCl 2 + H 2 If you are given 70 g Zn and 71 g HCl, what will be the mass of the salt produced? HCl is our limiting reactant 2 71 g HCl X 1 mol HCl X 1 mol ZnCl 2 X 136.3 g ZnCl 2 = 132.57 g ZnCl 2 36.5g HCl 2 mol HCl 1 mol ZnCl 2 132.57 g is the theoretical yield for this reaction.

11. % yield  % yield = experimental yield x 100% theoretical yield % expressing what you u collected vs what you were supposed to collect.

12. % Yield  If you did the reaction on the last two slides and you only collected 110 g of ZnCl 2, what is your % yield? % Yield = mass collected X 100% mass expected % Yield = 110 g X 100% = 82.97% 132.57 g

13.  Mg + HCl -- > MgCl 2 + H 2 If you have 25.4 g Mg and combine it with 50.0 mL of 1.25 M HCl, what mass of MgCl2 will you collect?

14. Do Now 9/22 – Do this then discuss questions from your reading on gasses 2  Zn + HCl  ZnCl 2 + H 2 If you are given 70.0 g Zn and 71.0 g HCl, what will be the mass of the salt produced? If you did the reaction and you only collected 110 g of ZnCl 2, what is your % yield?

15. Do Now 9/24 – Do this then discuss questions from your reading on gasses A sample of a compound contains 0.99 g Na, 1.365 g S and 1.021 g O. Determine its empirical formula. ATP is an important molecule in living cells. A sample with a mass of 0.8138 g was analyzed and found to contain 0.1927 g C, 0.02590 g H, 0.1124 g N, and 0.1491 g P. The remainder was O. Determine the empirical formula of ATP. Its formula mass was found to be 507 g mol-1. Determine the molecular formula.

16. Do Now 9/25 A sample of a compound contains 0.99 g Na, 1.365 g S and 1.021 g O. Determine its empirical formula. ATP is an important molecule in living cells. A sample with a mass of 0.8138 g was analyzed and found to contain 0.1927 g C, 0.02590 g H, 0.1124 g N, and 0.1491 g P. The remainder was O. Determine the empirical formula of ATP. Its formula mass was found to be 507 g mol-1. Determine the molecular formula.