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Difference between revisions of "Yield (Chemistry)"

(Examples)
(Examples)
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===Examples===
 
===Examples===
 
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Calculate the '''yield''' of [[Magnesium Oxide]] given 48g of [[Magnesium]] and excess [[Oxygen]].
 
Calculate the '''yield''' of [[Magnesium Oxide]] given 48g of [[Magnesium]] and excess [[Oxygen]].
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Find the '''yield''' of [[Water]] when 320g of [[Methane]] [[Chemical reaction|reacts]] with excess [[Oxygen]].
 
Find the '''yield''' of [[Water]] when 320g of [[Methane]] [[Chemical reaction|reacts]] with excess [[Oxygen]].
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Find the '''yield''' of [[Sodium Chloride]] when 7.3 of [[Hydrochloric Acid]] [[Chemical Reaction|reacts]] with excess [[Sodium Hydroxide]].
 
Find the '''yield''' of [[Sodium Chloride]] when 7.3 of [[Hydrochloric Acid]] [[Chemical Reaction|reacts]] with excess [[Sodium Hydroxide]].
 
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'''Write the Balanced Symbol Equation'''
 
'''Write the Balanced Symbol Equation'''
  
 
2Mg + O<sub>2</sub> → 2MgO
 
2Mg + O<sub>2</sub> → 2MgO
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'''Write the Balanced Symbol Equation'''
 
'''Write the Balanced Symbol Equation'''
  
 
CH<sub>4</sub> + 2O<sub>2</sub> → 2H<sub>2</sub>O + CO<sub>2</sub>
 
CH<sub>4</sub> + 2O<sub>2</sub> → 2H<sub>2</sub>O + CO<sub>2</sub>
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'''Write the Balanced Symbol Equation'''
 
'''Write the Balanced Symbol Equation'''
  
 
NaOH + HCl → NaCl + H<sub>2</sub>O
 
NaOH + HCl → NaCl + H<sub>2</sub>O
 
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Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]].
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'''Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]].'''
  
 
M<sub>r</sub> of Mg = 24g
 
M<sub>r</sub> of Mg = 24g
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M<sub>r</sub> of MgO = 40g
 
M<sub>r</sub> of MgO = 40g
  
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Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]].
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'''Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]].'''
  
 
M<sub>r</sub> of CH<sub>4</sub> = 16g
 
M<sub>r</sub> of CH<sub>4</sub> = 16g
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M<sub>r</sub> of H<sub>2</sub>O = 18g
 
M<sub>r</sub> of H<sub>2</sub>O = 18g
  
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Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]].
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'''Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]].'''
  
 
M<sub>r</sub> of NaOH = 40g
 
M<sub>r</sub> of NaOH = 40g
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M<sub>r</sub> of NaCl = 58.5g
 
M<sub>r</sub> of NaCl = 58.5g
 
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State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
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'''State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
  
 
2 [[mole]]s of Mg are needed for every 2 [[mole]]s of MgO
 
2 [[mole]]s of Mg are needed for every 2 [[mole]]s of MgO
  
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State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
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'''State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
  
 
1 [[mole]] of CH<sub>4</sub> is needed for every 2 [[mole]]s H<sub>2</sub>O
 
1 [[mole]] of CH<sub>4</sub> is needed for every 2 [[mole]]s H<sub>2</sub>O
  
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State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
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'''State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
  
 
1 [[mole]] of HCl and NaOH are needed for every 1 [[mole]] of NaCl
 
1 [[mole]] of HCl and NaOH are needed for every 1 [[mole]] of NaCl
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Find the number of [[mole]]s supplied of the known [[mass]].
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'''Find the number of [[mole]]s supplied of the known [[mass]].'''
  
 
No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>
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'''Yield''' = 80g
 
'''Yield''' = 80g
  
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Find the number of [[mole]]s supplied of the known [[mass]].
+
'''Find the number of [[mole]]s supplied of the known [[mass]].'''
  
 
No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>
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'''Yield''' = 720g
 
'''Yield''' = 720g
  
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| style="height:20px; width:300px; text-align:left;" |
Find the number of [[mole]]s supplied of the known [[mass]].
+
'''Find the number of [[mole]]s supplied of the known [[mass]].'''
  
 
No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>

Revision as of 17:51, 12 January 2019

Key Stage 4

Meaning

Yield is the mass of product in a chemical reaction.

About Yield

The theoretical yield of product form a chemical reaction can be calculated using the balanced symbol equation and the relative formula masses of all the molecules involved.
The experimental yield from a chemical reaction is often not as large as the theoretical yield because not all of the reactants will react. This experimental yield is often expressed as a percentage of the theoretical yield and is known as the Percentage Yield.

Examples

Calculate the yield of Magnesium Oxide given 48g of Magnesium and excess Oxygen.

Find the yield of Water when 320g of Methane reacts with excess Oxygen.

Find the yield of Sodium Chloride when 7.3 of Hydrochloric Acid reacts with excess Sodium Hydroxide.

Write the Balanced Symbol Equation

2Mg + O2 → 2MgO

Write the Balanced Symbol Equation

CH4 + 2O2 → 2H2O + CO2

Write the Balanced Symbol Equation

NaOH + HCl → NaCl + H2O

Find the Relative Formula Mass of the relevant reactants and product.

Mr of Mg = 24g

Mr of MgO = 40g

Find the Relative Formula Mass of the relevant reactants and product.

Mr of CH4 = 16g

Mr of H2O = 18g

Find the Relative Formula Mass of the relevant reactants and product.

Mr of NaOH = 40g

Mr of HCl = 36.5g

Mr of NaCl = 58.5g

State the ratio of moles of each chemical needed.

2 moles of Mg are needed for every 2 moles of MgO

State the ratio of moles of each chemical needed.

1 mole of CH4 is needed for every 2 moles H2O

State the ratio of moles of each chemical needed.

1 mole of HCl and NaOH are needed for every 1 mole of NaCl

Find the number of moles supplied of the known mass.

No. Moles = \(\frac{Mass}{M_r}\)

No. Moles = \(\frac{48}{24}\)

No. Moles = 2 mol

Therefore 2 moles of MgO is produced.

2 moles of MgO = 40x2

Yield = 80g

Find the number of moles supplied of the known mass.

No. Moles = \(\frac{Mass}{M_r}\)

No. Moles = \(\frac{320}{16}\)

No. Moles = 20 mol

Therefore 40 moles of H2O are produced.

40 moles of CO2 = 18x40

Yield = 720g

Find the number of moles supplied of the known mass.

No. Moles = \(\frac{Mass}{M_r}\)

No. Moles = \(\frac{7.3}{36.5}\)

No. Moles = 0.2 mol

Therefore 0.2 mole of Sodium Chloride is produced.

0.2 mole of NaCl = 11.7g

Yield = 11.7g