MAYHAN Ch. 10/12 Moles to % yield Review ws DEFINE MOLE. MAYHAN HOW IS THE MOLE USED IN CHEMISTRY? A mole is 6.02 x 10 23 “things” These things are any.

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

DEFINE MOLE. MAYHAN HOW IS THE MOLE USED IN CHEMISTRY? A mole is 6.02 x “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

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 grams of O = grams of O 2 O2O2 O= g X 2 = g 1 mole = grams = 3.07 X 10 2 grams of O 2 MAYHAN

Moles to grams conversion moles iron III nitrate Place starting amount here mole Fe(NO 3 ) 3 Place conversion here 1 mole Fe(NO 3 ) grams of Fe(NO 3 ) = grams of Fe(NO 3 ) 3 Fe(NO 3 ) 3 Fe= g X 1 = g O= g X 9 = g N= g X 3 = g 1 mole = grams = 8.88 x10 1 grams of Fe(NO 3 ) 3 MAYHAN

Grams to moles conversions 6.25 g Ammonium sulfate 6.25 g of (NH 4 ) 2 SO 4 Place starting amount here mole of (NH 4 ) 2 SO g of (NH 4 ) 2 SO = mole of (NH 4 ) 2 SO 4 (NH 4 ) 2 SO 4 N= g X 2 = g H= g X 8 = 8.06 g S= g X 1 = g O= g X 4 = g 1 mole = grams = x mole of (NH 4 ) 2 SO 4 MAYHAN

Grams to moles conversions 235 g potassium carbonate 235 g of K 2 CO 3 Place starting amount here mole of K 2 CO g of K 2 CO = mole of K 2 CO 3 K 2 CO 3 K= g X 2 = g C= g X 1 = g O= g X 3 = g 1 mole = grams = 1.70 mole of K 2 CO 3 MAYHAN

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) g H 2 O 78.00g Al(OH) 3 1 mole Al(OH) 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 g HCl 1 mole HCl 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

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) g HCl 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

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%

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

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 g – 27.25g = 2.60 g

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 g KI 3.61 g PbI g KI 1 mole KI461 g PbI 2 1 mol PbI 2 2 mole KI 1 mol PbI

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 g 3.10 g Pb(NO 3 ) g PbI g Pb(NO 3 ) 2 1 mole Pb(NO 3 ) g PbI 2 1 mol PbI 2 1 mole Pb(NO 3 ) 2 1 mol PbI

MAYHAN Limited Reactan t Theoretical yield Excess Reactant 2.60 g KI 3.61 g PbI g KI 1 mole KI461 g PbI 2 1 mol PbI 2 2 mole KI 1 mol PbI g Pb(NO 3 ) g PbI g Pb(NO 3 ) 2 1 mole Pb(NO 3 ) 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?________________________

MAYHAN Limited Reactan t 2.60 g KI 2.59 g Pb(NO 3 ) g KI 1 mole KI 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 grams wasted!!

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

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