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Highland Science Department Percentage YieldPowerPoint Presentation

Highland Science Department Percentage Yield

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e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.

e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.

e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.

e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.

e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.

Percentage Yield

Percentage Yield

Percentage Yield: a comparison of the mass of product actually obtained in an experiment to the amount theoretically possible.

Percentage Yield

Percentage Yield: a comparison of the mass of product actually obtained in an experiment to the amount theoretically possible.

-it's rare to produce the same amount of product as predicted by the balanced chemical equation

Percentage Yield

Percentage Yield: a comparison of the mass of product actually obtained in an experiment to the amount theoretically possible.

-it's rare to produce the same amount of product as predicted by the balanced chemical equation

Reasons: -side reactions

-reaction does not go to completion

-loss of product during separation

Percentage Yield

e.g. 1. If 189 g of lead (II) sulfide was actually obtained in a reaction for which the theoretical yield was 239 g, calculate the percentage yield.

Percentage Yield

e.g. 1. If 189 g of lead (II) sulfide was actually obtained in a reaction for which the theoretical yield was 239 g, calculate the percentage yield.

G: actual yield of PbS = 189 g

theoretical yield of PbS = 239 g

U: % yield

S: % yield = actual yield x 100%

theoretical yield

Percentage Yield

e.g. 1. If 189 g of lead (II) sulfide was actually obtained in a reaction for which the theoretical yield was 239 g, calculate the percentage yield.

G: actual yield of PbS = 189 g

theoretical yield of PbS = 239 g

U: % yield

S: % yield = actual yield x 100%

theoretical yield

S: % yield = 189 g x 100%

239 g

= 79.1 %

Percentage Yield

e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.

Percentage Yield

e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.

G: 2Na(s) + Cl2(g) 2NaCl(s)MM Cl2

mass of Na = 8.30 g Cl = 2 x 35.4 = 70.8 g

mass of Cl2 = 19.5 g MM NaCl

actual yield of NaCl = 19.5 g Na = 1 x 23.0 = 23.0

Cl = 1 x 35.4 = 35.4

58.4 g

Percentage Yield

e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.

G: 2Na(s) + Cl2(g) 2NaCl(s)M.M. Cl2

mass of Na = 8.30 g Cl = 2 x 35.4 = 70.8 g

mass of Cl2 = 19.5 g M.M. NaCl

actual yield of NaCl = 19.5 g Na = 1 x 23.0 = 23.0

Cl = 1 x 35.4 = 35.4

U: percentage yield 58.4 g

S: 1: limiting reactant = mass ÷ lowest #

M.M.

2: limiting limiting NaCl NaCl

reactant reactant moles mass

mass moles

3: % yield = actual mass x 100%

theoretical mass

Percentage Yield

S: Na amount = 8.30g = 0.361 mol

Cl2 amount = 14.0g = 0.198 mol

Percentage Yield

S: Na amount = 8.30g = 0.361 mol = 1.82

0.198 mol

Cl2 amount = 14.0g = 0.198 mol = 1

0.198 mol

Percentage Yield

S: Na amount = 8.30g = 0.361 mol = 1.82 < 2

0.198 mol limiting

Cl2 amount = 14.0g = 0.198 mol = 1 = 1 excess

0.198 mol

Percentage Yield

S: Na amount = 8.30g = 0.361 mol = 1.82 < 2

0.198 mol limiting

Cl2 amount = 14.0g = 0.198 mol = 1 = 1 excess

0.198 mol

theoretical = 8.30 g Na

mass NaCl

= 8.30 x 58.4 g NaCl

23.0

= 21.1 g

Percentage Yield

S: Na amount = 8.30g = 0.361 mol = 1.82 < 2

0.198 mol limiting

Cl2 amount = 14.0g = 0.198 mol = 1 = 1 excess

0.198 mol

theoretical = 8.30 g Na

mass NaCl

= 8.30 x 58.4 g NaCl

23.0

= 21.1 g

% yield = 19.5 g x 100%

21.1 g

= 92.4 %

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