Presentation is loading. Please wait.

Presentation is loading. Please wait.

Lecture 129/28/05 So what have you learned in the last few days?

Similar presentations


Presentation on theme: "Lecture 129/28/05 So what have you learned in the last few days?"— Presentation transcript:

1 Lecture 129/28/05 So what have you learned in the last few days?

2 Skill 1: Balancing Equations Fe 2 O 3 (s) + CO(g)  Fe(s) + CO 2 (g)

3 Skill 1: Balancing Equations Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g)

4 Skill 2: Using Stoichiometry Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g) 1 mole Fe 2 O 3 3 mole CO 2 mole Fe 1 mole Fe 2 O 3 3 mole CO 2 If you have 1 mole of Fe 2 O 3

5 Skill 2: Using Stoichiometry Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g) 0.5 mole Fe 2 O 3 1 mole Fe 1.5 mole CO 1.5 mole CO 2 1 mole Fe If you have 1 mole of Fe

6 How much CO does it take to react with 100 g of Fe 2 O 3 ? Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g) 0.626 mole Fe 2 O 3 1.88 mole CO 100 g Fe 2 O 3 52.6 g CO

7 How much Fe can be formed if you have 100 grams of CO and 100 grams of Fe 2 O 3 ? Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g) 0.626 mole Fe 2 O 3 3.57 mole CO 100 g Fe 2 O 3 100 g CO Find limiting reagent

8 How much Fe can be formed if you have 100 grams of CO and 100 grams of Fe 2 O 3 ? Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g) Limiting reactant method 1: StoichiometricActual < Fe 2 O 3 is limiting reagent

9 How much Fe can be formed if you have 100 grams of CO and 100 grams of Fe 2 O 3 ? Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g) 0.626 mole Fe 2 O 3 3.57 mole CO 1.25 mole Fe 2.38 mole Fe Limiting reactant method 2: Fe 2 O 3 is limiting reagent

10 How much Fe can be formed if you have 100 grams of CO and 100 grams of Fe 2 O 3 ? Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g) 0.626 mole Fe 2 O 3 1.25 mole Fe Once you know limiting reactant, use stoichiometry 69.8 g Fe 100 g Fe 2 O 3

11 How much Fe can be formed if you have 100 grams of CO and 100 grams of Fe 2 O 3 ? If you form 50 g of Fe, what is the percent yield? Fe 2 O 3 (s) + 3 CO(g)  2 Fe(s) + 3 CO 2 (g) 0.626 mole Fe 2 O 3 1.25 mole Fe 69.8 g Fe100 g Fe 2 O 3

12 Aluminum reacts with oxygen to give aluminum oxide. 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) What amount of O 2 (in moles) is needed for complete reaction with 6.0 mol of Al? What mass of Al 2 O 3, grams, can be produced?

13 Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride. CaO(s) + 2NH 4 Cl(s)  2NH 3 (g) + H 2 O(g) + CaCl 2 (s) If 112 g of CaO and 224 g of NH 4 Cl are mixed, a) What mass of NH 3 can be produced? b) What mass of the excess reactant remains after the ammonia has been formed?

14 The deep blue compound Cu(NH 3 ) 4 SO 4 is made by the reaction of copper (II) sulfate and ammonia. CuSO 4 (aq) + 4NH 3 (aq)  Cu(NH 3 ) 4 SO 4 (aq) a) If you use 10.0 g of CuSO 4 and excess NH 3, what is the theoretical yield of Cu(NH 3 ) 4 SO 4 ? b) If you obtain 12.6g of Cu(NH 3 ) 4 SO 4, what is the percent yield?


Download ppt "Lecture 129/28/05 So what have you learned in the last few days?"

Similar presentations


Ads by Google