Theoretical Yield The amount of product(s) calculated using stoichiometry problems.

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Presentation transcript:

Theoretical Yield The amount of product(s) calculated using stoichiometry problems

Actual Yield The amount of product actually recovered when the reaction is done in the lab

Percent Yield (Actual yield/Theoretical Yield) x 100

A reaction between solid sulfur and oxygen produces sulfur dioxide. 1S 6 (s) + 6O 2 (g)  6SO 2 (g)

The reaction started with 384 grams of S 6 (s). Assume an unlimited supply of oxygen. What is the theoretical (predicted) yield and the percent yield if only 680 grams of sulfur dioxide are produced? Find moles of S g S 6 ( 1 moles S 6 / 192 g S 6 ) = 2 moles S 6 Find moles of SO 2 2 moles S 6 (6 moles SO 2 /1 mole S 6 ) = 12 moles SO 2

Find theoretical grams SO 2 12 moles SO2 (64 g SO2/1mole) = 768 g SO2 Find Percent Yield (680 g / 768g) x100 = 89%

Zinc reacts with copper sulfate in a single replacement reaction as follows Zn (s) + CuSO 4 (aq)  ZnSO 4 (aq) + Cu (s) grams of zinc metal were added to excess copper sulfate dissolved in a water solution grams of copper were recovered. Calculate the theoretical yield of copper in this experiment

Step 1: Find the theoretical mass of Cu Mass of Cu = – 50.00g Zn(mol/65.38gZn) = mol Zn – mol Zn (1 molCu/1 mol Zn) = mol Cu – mol Cu (63.55g Cu/mol) = g Cu

Step 2: Find Percent Yield % Yield = (42.50 g Cu/48.60 g Cu)x 100 = 87.44%

Silver can produced by reacting silver nitrate with magnesium in the following reaction Mg(s) + 2 AgNO 3 (aq)  Mg(NO 3 ) 2 (aq) + 2 Ag (s)

How much Silver can be recovered by reacting a silver nitrate solution with grams of powdered magnesium g Mg(mol/24.31 g Mg) = mol Mg mol Mg ( 2mol Ag/ 1 mol Mg) = mol Ag mol Ag (107.9 g/mol) =443.8 g Ag

Assume that 95% of the silver can be recovered g theoretically produced g (0.95) = g Ag actually produced

MOLARITY Concentration

Solutions Solute: the substance being dissolved Solvent: the substance doing the dissolving

Units of Concentration Molarity – Ratio of – “Amount” of Solute/ “Amount” of Solution

Molarity Moles of solute/L of solution M

Prepare 500mL of a 1M sucrose solution

Prepare 250 mL of a 6M solution of dye 4 drops = 1 mole