Difference between revisions of "Yield (Chemistry)"
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===About Yield=== | ===About Yield=== | ||
− | : The theoretical '''yield''' of [[product]] | + | : The theoretical '''yield''' of [[product]] from a [[Chemical Reaction|chemical reaction]] can be calculated using the [[Balanced Symbol Equation|balanced symbol equation]] and the [[Relative Formula Mass|relative formula masses]] of all the [[molecule]]s involved. |
: The experimental '''yield''' from a [[Chemical Reaction|chemical reaction]] is often not as large as the theoretical '''yield''' because not all of the [[reactant]]s will [[Chemical Reaction|react]]. This experimental '''yield''' is often expressed as a [[percentage]] of the theoretical '''yield''' and is known as the [[Percentage Yield]]. | : The experimental '''yield''' from a [[Chemical Reaction|chemical reaction]] is often not as large as the theoretical '''yield''' because not all of the [[reactant]]s will [[Chemical Reaction|react]]. This experimental '''yield''' is often expressed as a [[percentage]] of the theoretical '''yield''' and is known as the [[Percentage Yield]]. | ||
===Examples=== | ===Examples=== | ||
{| class="wikitable" | {| class="wikitable" | ||
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" | |
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 [[ | + | Find the '''yield''' of [[Water]] when 320g of [[Methane]] [[Chemical reaction|reacts]] with excess [[Oxygen]]. |
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" | |
− | Find the '''yield''' of [[Sodium Chloride]] when | + | Find the '''yield''' of [[Sodium Chloride]] when 7.3 of [[Hydrochloric Acid]] [[Chemical Reaction|reacts]] with excess [[Sodium Hydroxide]]. |
|- | |- | ||
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" | |
'''Write the Balanced Symbol Equation''' | '''Write the Balanced Symbol Equation''' | ||
2Mg + O<sub>2</sub> → 2MgO | 2Mg + O<sub>2</sub> → 2MgO | ||
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" | |
'''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> | ||
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:center;" | |
'''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 | ||
|- | |- | ||
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" | |
− | Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]]. | + | '''Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]].''' |
− | M<sub>r</sub> of | + | M<sub>r</sub> of Mg = 24g |
− | M<sub>r</sub> of | + | M<sub>r</sub> of MgO = 40g |
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" | |
− | Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]]. | + | '''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 | ||
− | M<sub>r</sub> of | + | M<sub>r</sub> of H<sub>2</sub>O = 18g |
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" | |
− | Find the [[Relative Formula Mass]] of the relevant [[reactant]]s and [[product]]. | + | '''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 | ||
|- | |- | ||
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" | |
− | State the [[ratio]] of [[mole]]s of each [[chemical]] needed. | + | '''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 | ||
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" | |
− | State the [[ratio]] of [[mole]]s of each [[chemical]] needed. | + | '''State the [[ratio]] of [[mole]]s of each [[chemical]] needed.''' |
− | 1 [[mole]] of CH<sub>4</sub> is needed for every | + | 1 [[mole]] of CH<sub>4</sub> is needed for every 2 [[mole]]s H<sub>2</sub>O |
− | | style="height:20px; width: | + | | style="height:20px; width:300px; text-align:left;" | |
− | State the [[ratio]] of [[mole]]s of each [[chemical]] needed. | + | '''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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|- | |- | ||
− | | style="height:20px; width: | + | | 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> | ||
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No. [[Mole]]s = <math>\frac{48}{24}</math> | No. [[Mole]]s = <math>\frac{48}{24}</math> | ||
− | No. [[Mole]]s = 2 | + | No. [[Mole]]s = 2 mol |
− | Therefore | + | Therefore 2 [[mole]]s of MgO is produced. |
− | + | 2 [[mole]]s of MgO = 40x2 | |
− | | style="height:20px; width: | + | '''Yield''' = 80g |
− | Find the number of [[mole]]s supplied of the known [[mass]]. | + | |
+ | | style="height:20px; width:300px; text-align:left;" | | ||
+ | '''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> | ||
− | No. [[Mole]]s = <math>\frac{ | + | No. [[Mole]]s = <math>\frac{320}{16}</math> |
+ | |||
+ | No. [[Mole]]s = 20 mol | ||
− | + | Therefore 40 [[mole]]s of H<sub>2</sub>O are produced. | |
− | + | 40 [[mole]]s of CO<sub>2</sub> = 18x40 | |
− | + | '''Yield''' = 720g | |
− | | style="height:20px; width: | + | | 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> | ||
− | No. [[Mole]]s = <math>\frac{ | + | No. [[Mole]]s = <math>\frac{7.3}{36.5}</math> |
− | No. [[Mole]]s = 0. | + | No. [[Mole]]s = 0.2 mol |
− | Therefore 0. | + | Therefore 0.2 [[mole]] of [[Sodium Chloride]] is produced. |
− | 0. | + | 0.2 [[mole]] of NaCl = 11.7g |
+ | '''Yield''' = 11.7g | ||
|} | |} | ||
+ | |||
+ | ===References=== | ||
+ | ====AQA==== | ||
+ | |||
+ | :[https://www.amazon.co.uk/gp/product/0008158754/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0008158754&linkCode=as2&tag=nrjc-21&linkId=27ad53b0283feeff7fc5ae04a9e205f621 ''Yield, page 66-7, GCSE Biology; Student Book, Collins, AQA ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/1471851346/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851346&linkCode=as2&tag=nrjc-21&linkId=3ac654f4b0da781c49c855a1af4c92ea ''Yield, page 78, GCSE Chemistry, Hodder, AQA ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/178294639X/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=178294639X&linkCode=as2&tag=nrjc-21&linkId=51599bb45a2bfaf7c1b6a978b2ca2616 ''Yield, pages 184-186, GCSE Combined Science Trilogy; Chemistry, CGP, AQA ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/0198359381/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0198359381&linkCode=as2&tag=nrjc-21&linkId=47c8d1ae58d8b3a5e2094cd447154558 ''Yield, pages 68-71, GCSE Chemistry; Third Edition, Oxford University Press, AQA ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/0008158762/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0008158762&linkCode=as2&tag=nrjc-21&linkId=a0fffa35b3ea49a63404f6704e0df7cc ''Yield, pages 97, 216-219, 317, 323, GCSE Chemistry; Student Book, Collins, AQA ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/1782945962/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945962&linkCode=as2&tag=nrjc-21&linkId=476bb5c8d1dfb5c08ac81b6d4d1c98d8 ''Yields, pages 141, 144, 216-218, GCSE Chemistry, CGP, AQA ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/1782945571/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945571&linkCode=as2&tag=nrjc-21&linkId=9e29fad914244909903e5e93f8a01d272 ''Yields, pages 45, 49, GCSE Chemistry; The Revision Guide, CGP, AQA ''] | ||
+ | |||
+ | ====Edexcel==== | ||
+ | |||
+ | :[https://www.amazon.co.uk/gp/product/1782948147/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782948147&linkCode=as2&tag=nrjc-21&linkId=f63dcd8345f4e49c717b39a228a36c7c ''Yields, pages 172-174, 189-191, 201-203, GCSE Chemistry, CGP, Edexcel ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/1782945725/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945725&linkCode=as2&tag=nrjc-21&linkId=694be7494de75af3349537d34e13f7f0 ''Yields, pages 66, 68, GCSE Chemistry; The Revision Guide, CGP, Edexcel ''] | ||
+ | |||
+ | ====OCR==== | ||
+ | :[https://www.amazon.co.uk/gp/product/0198359829/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0198359829&linkCode=as2&tag=nrjc-21&linkId=90e8d7b4f039d53035238fa0320fe00b ''Yield, pages 158-161, 190, 200-203, 205, Gateway GCSE Chemistry, Oxford, OCR ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/1782945679/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945679&linkCode=as2&tag=nrjc-21&linkId=a2db42f7b4bdf10cafaafa3bb9120940 ''Yield, pages 66, 74, 80, 82, Gateway GCSE Chemistry; The Revision Guide, CGP, OCR ''] | ||
+ | :[https://www.amazon.co.uk/gp/product/1782945660/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945660&linkCode=as2&tag=nrjc-21&linkId=83aa4500ad7759e7f401a1c5ba5df758 ''Yields (in agriculture), pages 86, 87, Gateway GCSE Biology; The Revision Guide, CGP, OCR ''] |
Latest revision as of 12:15, 25 December 2019
Contents
Key Stage 4
Meaning
Yield is the mass of product in a chemical reaction.
About Yield
- The theoretical yield of product from 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 |
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 |
References
AQA
- Yield, page 66-7, GCSE Biology; Student Book, Collins, AQA
- Yield, page 78, GCSE Chemistry, Hodder, AQA
- Yield, pages 184-186, GCSE Combined Science Trilogy; Chemistry, CGP, AQA
- Yield, pages 68-71, GCSE Chemistry; Third Edition, Oxford University Press, AQA
- Yield, pages 97, 216-219, 317, 323, GCSE Chemistry; Student Book, Collins, AQA
- Yields, pages 141, 144, 216-218, GCSE Chemistry, CGP, AQA
- Yields, pages 45, 49, GCSE Chemistry; The Revision Guide, CGP, AQA
Edexcel
- Yields, pages 172-174, 189-191, 201-203, GCSE Chemistry, CGP, Edexcel
- Yields, pages 66, 68, GCSE Chemistry; The Revision Guide, CGP, Edexcel