2. MAYHAN Ch. 10/12 Moles to % yield Review ws

3. DEFINE MOLE. MAYHAN HOW IS THE MOLE USED IN CHEMISTRY? A mole is 6.02 x 10 23 “things” These things are any tiny tiny things When balancing equations, the coefficients represent a “packaged” number of tiny things (mole). This can be converted into grams through molar mass

4. Moles to grams conversion 9.59 mol of O 2 Place starting amount here 9.59 mole O 2 Place conversion here 1 mole O 2 ----------31.998 grams of O 2 -------------- = 306.8 grams of O 2 O2O2 O= 15.999 g X 2 = 31.998 g 1 mole = 31.998 grams = 3.07 X 10 2 grams of O 2 MAYHAN

5. Moles to grams conversion 0.367 moles iron III nitrate Place starting amount here 0.367 mole Fe(NO 3 ) 3 Place conversion here 1 mole Fe(NO 3 ) 3 --------------- 241.85 grams of Fe(NO 3 ) 3 -------------- = 88.75 grams of Fe(NO 3 ) 3 Fe(NO 3 ) 3 Fe= 55.845 g X 1 = 55.845 g O= 15.999 g X 9 = 143.99 g N= 14.0067 g X 3 = 42.02 g 1 mole = 241.85 grams = 8.88 x10 1 grams of Fe(NO 3 ) 3 MAYHAN

6. Grams to moles conversions 6.25 g Ammonium sulfate 6.25 g of (NH 4 ) 2 SO 4 Place starting amount here ---------- 1 mole of (NH 4 ) 2 SO 4 132.135 g of (NH 4 ) 2 SO 4 ----------- = 0.0473 mole of (NH 4 ) 2 SO 4 (NH 4 ) 2 SO 4 N= 14.007 g X 2 = 28.014 g H= 1.0079 g X 8 = 8.06 g S= 32.065 g X 1 = 32.065 g O= 15.999 g X 4 = 63.996 g 1 mole = 132.135 grams = 0.0473 x 10 -2 mole of (NH 4 ) 2 SO 4 MAYHAN

7. Grams to moles conversions 235 g potassium carbonate 235 g of K 2 CO 3 Place starting amount here ---------- 1 mole of K 2 CO 3 138.2 g of K 2 CO 3 ----------- = 1.7004 mole of K 2 CO 3 K 2 CO 3 K= 39.098 g X 2 = 78.196 g C= 12.01 g X 1 = 12.01 g O= 15.999 g X 3 = 47.997 g 1 mole = 138.2 grams = 1.70 mole of K 2 CO 3 MAYHAN

8. 2. In a double replacement reaction, 3.50 g Aluminum hydroxide reacts 7.00 grams of hydrochloridic acid (hydrogen chloride) to create 2 products. Using the water product, identify the limiting reactant, the amount of theoretical yield and how much excess was produced. Balance equation here: __Al(OH) 3 + __HCl  __AlCl 3 + __H 2 O 3.50 g Al(OH) 3 2.42 g H 2 O 78.00g Al(OH) 3 1 mole Al(OH) 3 18.00 g H 2 O 1 mol H 2 O 1 mole Al(OH) 3 3 mol H 2 O 7.00 g HCl 3.46 g H 2 O 36.46 g HCl 1 mole HCl 18.00 g H 2 O 1 mole H 2 O 3 mole HCl 3 mole H 2 O Which REACTANT is the limiting reactant? ( Al(OH) 3 or HCl )? Limited Reactan t 33 Theoretical yield Excess Reactant

9. 2. In a double replacement reaction, 3.50 g Aluminum hydroxide reacts 7.00 grams of hydrochloride acid (hydrogen chloride) to create 2 products. Using the water product, identify the limiting reactant, the amount of theoretical yield and how much excess was produced. Balance equation here: Start with the limited reactant!! Used In reaction 7.00 grams HCl minus 4.91 g HCl = Given Used In reaction 2.09 g EXCESS 3.50 g Al(OH) 3 4.91 g HCl 36.46 g HCl 1 mole HCl78.0 g Al(OH) 3 1 mole Al(OH) 3 3 mole HCl 1 mole Al(OH) 3 __Al(OH) 3 + __HCl  __AlCl 3 + __H 2 O 33

10. A student did a lab and retrieved 2.20 grams of water, what is the percent yield? Theoretically we got AT BEST 2.42 grams of water 2.20 g water (actual) 2.42 g water (theo yield) X 100 = 90.9%

11. MAYHAN A student carried out a reaction between aqueous solution of lead II nitrate and potassium iodide. A double displacement reaction occured when the solutions were poured together and filtered, a bright yellow precipitate of lead II iodide collected on the filter paper. The following data was collected during the experiment: Pb(NO 3 ) 2 + KI  PbI 2 + KNO 3

12. Write the balance equation. 1. What is the mass of lead II nitrate used in the LAB EXPERIMENT? SHOW WORK AND LABEL!! 2. What is the mass of potassium iodide used in the LAB EXPERIMENT? SHOW WORK AND LABEL!! Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3 Mass of beaker – mass of beaker with lead II nitrate = lead II nitrate Mass of beaker – mass of beaker with potassium iodide = potassium iodide 30.35g – 27.25g = 3.10 g 29.85 g – 27.25g = 2.60 g

13. Write the balance equation. 3. Use the mass of your potassium iodide found in lab, what would be the amount of lead II iodide produced using this amount of potassium iodide? (2 points) SHOW WORK AND LABEL!! Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3 2.60 g KI 3.61 g PbI 2 165.998g KI 1 mole KI461 g PbI 2 1 mol PbI 2 2 mole KI 1 mol PbI 2 1198.6 331.996

14. Write the balance equation. 4. Use the mass of your lead II nitrate found in lab, what would be the amount of lead II iodide produced using this amount of lead II nitrate? (2 points) SHOW WORK AND LABEL!! Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3 3.10 g 3.10 g Pb(NO 3 ) 2 4.31 g PbI 2 331.208g Pb(NO 3 ) 2 1 mole Pb(NO 3 ) 2 461 g PbI 2 1 mol PbI 2 1 mole Pb(NO 3 ) 2 1 mol PbI 2 1429.1 331.996

15. MAYHAN Limited Reactan t Theoretical yield Excess Reactant 2.60 g KI 3.61 g PbI 2 165.998g KI 1 mole KI461 g PbI 2 1 mol PbI 2 2 mole KI 1 mol PbI 2 3.10 g Pb(NO 3 ) 2 4.31 g PbI 2 331.208g Pb(NO 3 ) 2 1 mole Pb(NO 3 ) 2 461 g PbI 2 1 mol PbI 2 1 mole Pb(NO 3 ) 2 1 mol PbI 2 5.What is the theoretical yield of lead II iodide?________________________

16. MAYHAN Limited Reactan t 2.60 g KI 2.59 g Pb(NO 3 ) 2 165.998g KI 1 mole KI 331.208 g Pb(NO 3 ) 2 1 mol Pb(NO 3 ) 2 2 mole KI 1 mol Pb(NO 3 ) 2 6. What is the amount of excess reactant? (2 points) SHOW WORK AND LABEL!! I started with 3.10 grams and I needed only 2.59 grams. Subtract to see how much was excess- 0.51 grams wasted!!

17. MAYHAN 7. What is the actual amount of lead II iodide produced in lab? mass of filter paper with lead II iodide - mass of filter paper 3.47g – 0.850g = 2.62 g

18. MAYHAN 8. What is the percent yield of this reaction? SHOW WORK AND LABEL!! Actual 2.62 g Theo 3.61 g X 100 = 72.6 g