180 likes | 471 Views
CHAPTER 9 – CHEMICAL QUANTITIES. INFORMATION FROM CHEMICAL EQUATIONS. 2H 2 + O 2 → 2H 2 O. 2 molecules 2 moles 0.84 moles 0.028 moles. 1 molecule 1 moles 0.42 moles 0.014 moles. 2 molecules 2 moles 0.84 moles 0.028 moles. 5B-1 (of 20).
E N D
CHAPTER 9 – CHEMICAL QUANTITIES INFORMATION FROM CHEMICAL EQUATIONS 2H2 + O2 → 2H2O 2 molecules 2 moles 0.84 moles 0.028 moles 1 molecule 1 moles 0.42 moles 0.014 moles 2 molecules 2 moles 0.84 moles 0.028 moles 5B-1 (of 20)
MASS CALCULATIONS Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O. C3H8O + O2→ 4½ CO2 + H2O 3 4 5B-2
MASS CALCULATIONS Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O. 9 C3H8O + O2→ CO2 + H2O 2 6 8 x g 5.00 g 2 mol 9 mol 2 mol C3H8O = 9 mol O2 5.00 g C3H8O x 1 mol C3H8O ___________________ 60.11 g C3H8O x 9 mol O2 _________________ 2 mol C3H8O x 32.00 g O2 ______________ 1 mol O2 = 12.0 g O2 5B-3
Calculate the mass of carbon dioxide produced from the 5.00 g of propanol. 9 C3H8O + O2→ CO2 + H2O 2 6 8 5.00 g x g 2 mol 6 mol 2 mol C3H8O = 6 mol CO2 5.00 g C3H8O x 1 mol C3H8O ___________________ 60.11 g C3H8O x 6 mol CO2 _________________ 2 mol C3H8O x 44.01 g CO2 ________________ 1 mol CO2 = 11.0 g CO2 5B-4
Calculate the mass of water produced from the 5.00 g of propanol. 5B-5
C3H8O + O2→ 2 9 6 CO2 + H2O 8 5.00 g 6.00 g 12.0 g 11.0 g 5B-6
Calculate the mass of oxygen needed to burn 25.0 g of propane, C3H8. C3H8 + O2→ 5 CO2 + H2O 3 4 25.0 g x g 1 mol 5 mol 1 mol C3H8 = 5 mol O2 5.00 g C3H8 x 1 mol C3H8 ________________ 44.17 g C3H8 x 5 mol O2 _______________ 1 mol C3H8 x 32.00 g O2 ______________ 1 mol O2 = 90.7 g O2 5B-7
Calculate the mass of AIR needed to burn the 25.0 g of propane, if air is 23.0% oxygen by mass. 100% air = 23.0% O2 100 g air = 23.0 g O2 90.7 g O2 x 100 g air _____________ 23.0 g O2 = 394 g air 5B-8
LIMITING REACTANT CALCULATIONS LIMITING REACTANT – The reactant that is completely used up in a reaction 5B-9
Calculate the mass of sulfur trioxide produced when 20.0 g of sulfur reacts with 25.0 g of oxygen. S + O2→ 2 3 2 SO3 25.0 g 20.0 g x g 3 mol 2 mol 2 mol 5B-10
Calculate the mass of sulfur trioxide produced when 20.0 g of sulfur reacts with 25.0 g of oxygen. S + O2→ 2 3 2 SO3 25.0 g 20.0 g x g 3 mol 2 mol 2 mol 20.0 g S x 1 mol S ____________ 32.07 g S x 2 mol SO3 ______________ 2 mol S x 64.07 g SO3 ______________ 1 mol SO3 = 49.9 g SO3 5B-11
Calculate the mass of sulfur trioxide produced when 20.0 g of sulfur reacts with 25.0 g of oxygen. S + O2→ 2 3 2 SO3 25.0 g 20.0 g x g 3 mol 2 mol 2 mol 20.0 g S x 1 mol S ____________ 32.07 g S x 2 mol SO3 ______________ 2 mol S x 64.07 g SO3 ______________ 1 mol SO3 = 49.9 g SO3 25.0 g O2 x 1 mol O2 _____________ 32.00 g O2 x 2 mol SO3 ______________ 3 mol O2 x 64.07 g SO3 ______________ 1 mol SO3 = 41.7 g SO3 O2 is the limiting reactant 41.7 g SO3 are produced 5B-12
Calculate the mass of aluminum sulfide produced when 9.00 g of aluminum reacts with 8.00 g of sulfur. 5B-13
PERCENT YIELD CALCULATIONS % Yield = collected amount of product x 100 _______________________________________ theoretical amount of product 5B-16
Calculate the percent yield if 5.00 g of magnesium metal reacts with excess silver nitrate to produce 40.5 g of silver metal. 2 Mg + AgCl → 2 Ag + MgCl2 5.00 g 40.5 g 5B-17
Calculate the percent yield if 5.00 g of magnesium metal reacts with excess silver nitrate to produce 40.5 g of silver metal. 2 Mg + AgCl → 2 Ag + MgCl2 5.00 g x g 1 mol 2 mol 5.00 g Mg x 1 mol Mg _______________ 24.31 g Mg x 2 mol Ag _____________ 1 mol Mg x 107.9 g Ag ______________ 1 mol Ag = 44.4 g Ag 40.5 g Ag actual 100 ___________________________ 44.4 g Ag theoretical = 91.2 % 5B-18
Calculate the percent yield if 7.50 g of hydrogen chloride reacts with excess zinc to produce 0.195 g of hydrogen gas. 5B-19
REVIEW FOR TEST Balance equations Indicate phases Predict products and balance Composition Decomposition Replacement Combustion Precipitation Acid-Base Mass calculations from chemical reactions Limiting reactant Percent yield 5B-20