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STOICHIOMETRY

STOICHIOMETRY. Determining Formulas. Using Stoichiometry to Determine a Formula. Burn 0.115 g of a hydrocarbon, C x H y , and produce 0.379 g of CO 2 and 0.1035 g of H 2 O . C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O What is the empirical formula of C x H y ?.

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STOICHIOMETRY

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  1. STOICHIOMETRY Determining Formulas

  2. Using Stoichiometry to Determine a Formula Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g of CO2 and 0.1035 g of H2O. CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O What is the empirical formula of CxHy?

  3. +O2 Puddle of CxHy +O2 0.115 g Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> CO2 + H2O 0.379 g0.1035 g First, recognize that all C in CO2 and all H in H2O is from CxHy. 0.379 g CO2 1 CO2 molecule forms for each C atom in CxHy 0.1035 g H2O 1 H2O molecule forms for each 2 H atoms in CxHy

  4. Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate amount of C in CO2 8.61 x 10-3 mol CO2 --> 8.61 x 10-3 mol C 2. Calculate amount of H in H2O 5.744 x 10-3 mol H2O -- >1.149 x 10-2 mol H

  5. Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O Now find ratio of mol H/mol C to find values of x and y in CxHy. 1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C3H4

  6. STOICHIOMETRY LIMITING REACTANTS

  7. Reactions Involving aLIMITING REACTANT • In a given reaction, there is not enough of one reagent to use up the other reagent completely. • The reagent in short supply LIMITS the quantity of product that can be formed.

  8. 2 NO(g) + O2 (g) 2 NO2(g) LIMITING REACTANTS Reactants Products Limiting reactant = ___________ Excess reactant = ____________

  9. Mass product Mass reactant Moles reactant Moles product STOICHIOMETRY CALCULATIONS Stoichiometric factor

  10. LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl2 + H2 Rxn 1: Zn remaining * More than enough Zn to use up the 0.100 mol HC Rxn 2: no Zn left * Not enough Zn to use up 0.100 mol HCl

  11. Mass product Mass reactant Moles reactant Moles product PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. What mass of AlCl3 can form? Coefficients Mole ratio

  12. Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

  13. Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio. 2 Al + 3 Cl2 ---> 2AlCl3 Reactants must be in the mole ratio

  14. Deciding on the Limiting Reactant 2 Al + 3 Cl2 ---> Al2Cl6 If There is not enough Al to use up all the Cl2 Lim. reagent = Al

  15. Deciding on the Limiting Reactant 2 Al + 3 Cl2 ---> 2AlCl3 If There is not enough Cl2 to use up all the Al Lim reagent = Cl2

  16. Step 2 of LR problem: Calculate moles of each reactant We have 5.40 g of Al and 8.10 g of Cl2

  17. This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Find mole ratio of reactants 2 Al + 3 Cl2 ---> Al2Cl6 Limiting reagent is Cl2

  18. grams Cl2 grams AlCl3 moles Cl2 moles AlCl3 2 Al + 3 Cl2 ---> 2AlCl3 Limiting reactant = Cl2 Base all calcs. on Cl2 Mix 5.40 g of Al with 8.10 g of Cl2. What mass of AlCl3 can form?

  19. CALCULATIONS: calculate mass of Al2Cl6 expected. Step 1: Calculate moles of AlCl3 expected based on LR. Step 2: Calculate mass of AlCl3 expected based on LR.

  20. How much of which reactant will remain when reaction is complete? • Cl2 was the limiting reactant. • Therefore, Al was present in excess. But how much? • First find how much Al was required. • Then find how much Al is in excess.

  21. Calculating Excess Al 2 Al + 3 Cl2 products 0.114 mol = LR 0.200 mol Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess

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