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Chemistry Ch 9 Warm-Up Practice quiz that should be written in your notes.

Chemistry Ch 9 Warm-Up Practice quiz that should be written in your notes. 4 moles of Li 2 CO 3. In the balanced Reaction: CO 2(g) + 2LiOH (s)  Li 2 CO 3(s) + H 2 O (l) How many moles of Li 2 CO 3 will be produced if 4 moles of CO 2 are reacted? What is the molar mass of LiOH ?.

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Chemistry Ch 9 Warm-Up Practice quiz that should be written in your notes.

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  1. Chemistry Ch 9 Warm-Up Practice quiz that should be written in your notes. 4 moles of Li2CO3 In the balanced Reaction: CO2(g) + 2LiOH(s) Li2CO3(s) + H2O(l) How many moles of Li2CO3 will be produced if 4 moles of CO2 are reacted? What is the molar mass of LiOH? 6.94g Li +16.00g O + 1.01g H = 23.95g LiOH 1 mol 1 mol 1 mol 1 mol

  2. Limiting Reactant = the reactant that limits the amount of the other reactants that can combine .

  3. Limiting Reactant • If 400 people want to get on a plane • And only 350 seats are available… • What is limiting the number of people on the flight? • There are 50 people in Excess. Extra The number of seats

  4. Excess Reactant = the substance that is NOT used up completely in a reaction. (Extra)

  5. SiO2(s) + 4HF (g) SiF4(g) + 2H2O (l) If 2.0 mol of HF are reacted with 4.5 mol of SiO2 , which is the limiting reagent?

  6. 4.5 mol of SiO2 2.0 mol of HF SiO2(s) + 4HF (g) SiF (g) + 2H2O (l) 2.0 mol HF x 1 mol SiO2=0.5 mol SiO2 4 mol HF The 2 mol of HF needed 0.5 mol of SiO2, but we had 4.5 mol of SiO2. (excess SiO2) Our limiting reagent was all used up.

  7. 2.00 mol of Zn 1.00 mol of S8 Zinc and sulfur react to form zinc sulfide. 8 Zn (s)+ S8(s)  8 ZnS (s) • If 2.00 mol of Zn are heated with 1.00 mol of S8, which is the limiting reactant? 2.00 mol Zn x 1 mol S8 = 0.25 mol S8 8 mol Zn is used up. Zn

  8. 2.00 mol of Zn 1.00 mol of S8 Zinc and sulfur react to form zinc sulfide. 8 Zn (s)+ S8(s)  8 ZnS (s) • If 2.00 mol of Zn are heated with 1.00 mol of S8, which is the limiting reactant? 1.00 mol S8x 8 mol Zn = 8.00 mol Zn 1 mol S8 is needed. Zn

  9. 2.00 mol of Zn 1.00 mol of S8 Zinc and sulfur react to form zinc sulfide. 8 Zn (s)+ S8(s)  8 ZnS (s) • How many mol of excess reactant remain? 2.00 mol Zn x 1 mol S8 = 0.25 mol S8 8 mol Zn is used up. 1.00 mol S8 - 0.25 mol S8 = 0.75 mol S8 remain.

  10. 2.00 mol of Zn 1.00 mol of S8 Zinc and sulfur react to form zinc sulfide. 8 Zn (s)+ S8(s)  8 ZnS (s) • How many mol of product are formed? 2.00 mol Zn x 8 mol ZnS = 2.00 mol ZnS 8 mol Zn

  11. Theoretical Yield = the maximum amount of product that can be produced from the supplied amount of reactant. • Actual Yield = the measured amount of product obtained from a reaction.

  12. Percent Yield = the ratio of actual yield to the theoretical yield x100% Percent yield = actual yield x 100% theoretical yield

  13. C6H6 (l) + Cl2 (g)  C6H5Cl (s) + HCl (g) When 36.8 g of C6H6 (l) react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percent yield of C6H5Cl?

  14. C6H6 (l) + Cl2 (g)  C6H5Cl (s) + HCl (g) Given: Reactant = 36.8 g of C6H6 (l) yield Product = 38.8 g ofC6H5Cl 1st What is the theoretical yield of C6H5Cl?

  15. 36.8 g of C6H6 38.8 g ofC6H5Cl C6H6 + Cl2 C6H5Cl + HCl molar ratio 1 mol C6H5Cl 1 mol C6H6 molar mass 6 C x 12.01 g/mol + 5 H x 1.01 g/mol + 1 Cl x 35.45 g/mol 112.56 g/mol C6H5Cl molar mass 6 C x 12.01 g/mol + 6 H x 1.01 g/mol 78.12 g/mol of C6H6

  16. Percent yield = actual yield x 100% theoretical yield Percent yield = 38.8g C6H5Clx 100%= 73.2% 53.0g C6H5Cl

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