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===Finding a Limiting Reactant===
====Limiting Reactants and Mass====
{| class="wikitable"
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| style="height:20px; width:250px; text-align:center;" |If there are are 6g of [[Carbon]] and 15g of [[Oxygen]] in the following [[Chemical Reaction|reaction]]:
No. [[Mole]]s = 0.5 mol
Therefore at a [[ratio]] of 1:1 then 0.5 [[mole]]s of O<sub>2</sub> is needed.
No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>
No. [[Mole]]s = 1 mol
Therefore at a [[ratio]] of 4:3 then 0.75 [[mole]]s of O<sub>2</sub> are needed.
No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>
No. [[Mole]]s = 1 mol
Therefore at a [[ratio]] of 1:6 then 6 [[mole]]s of O<sub>2</sub> are needed.
No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>
|}
====Limiting Reactants and Concentration====
{| class="wikitable"
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| style="height:20px; width:250px; text-align:center;" |If 0.1 dm<sup>3</sup> of 0.5[[Molarity|M]] [[Hydrochloric Acid]] is added to 0.1 dm<sup>3</sup> of 0.6[[Molarity|M]] [[Sodium Hydroxide]] which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?
| style="height:20px; width:250px; text-align:center;" |
'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>HCl+NaOH->NaCl+H2O</chem>
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'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>H2SO4+2NaOH->Na2SO4+2H2O</chem>
| style="height:20px; width:250px; text-align:center;" |
'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>2HCl+Mg(OH)2->MgCl2+2H2O</chem>
|-
| style="height:20px; width:250px; text-align:center;" |
Moles of HCl = 0.05 mol
Therefore at a [[ratio]] of 1:1 then 0.05 mol of NaOH are needed.
'' '''concentration of NaOH''' '' = <math>\frac{Moles (mol)}{volume (dm^3)}</math>
Moles of H<sub>2</sub>SO<sub>4</sub> = 0.06 mol
Therefore at a [[ratio]] of 1:2 then 0.12 mol of NaOH are needed.
'' '''concentration of NaOH''' '' = <math>\frac{Moles (mol)}{volume (dm^3)}</math>
Moles of HCl</sub> = 0.15 mol
Therefore at a [[ratio]] of 2:1 then 0.075 mol of Mg(OH)<sub>2</sub> are needed.
'' '''concentration of Mg(OH)<sub>2</sub>''' '' = <math>\frac{Moles (mol)}{volume (dm^3)}</math>
There is not enough Mg(OH)<sub>2</sub>, therefore Mg(OH)<sub>2</sub> is the '''limiting reactant'''.
|}
====Limiting Reactants and Volume====
{| class="wikitable"
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| style="height:20px; width:250px; text-align:center;" |If 12 dm<sup>3</sup> of [[Hydrogen]] [[gas]] [[Chemical Reaction|reacts]] with 15 dm<sup>3</sup> [[Fluorine]] [[gas]] which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?
| style="height:20px; width:250px; text-align:center;" |If 18 dm<sup>3</sup> of [[Hydrogen]] [[gas]] [[Chemical Reaction|reacts]] with 8 dm<sup>3</sup> [[Oxygen]] [[gas]] which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?
| style="height:20px; width:250px; text-align:center;" |If 9 dm<sup>3</sup> of [[Ethane]] [[Chemical Reaction|reacts]] with 32 dm<sup>3</sup> of [[Oxygen]] which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?
|-
| style="height:20px; width:250px; text-align:center;" |
'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>H2 + F2->2HF</chem>
| style="height:20px; width:250px; text-align:center;" |
'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>2H2 + O2-> 2H2O</chem>
| style="height:20px; width:250px; text-align:center;" |
'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>2C2H6 + 7O2->4CO2 + 6H2O</chem>
|-
| style="height:20px; width:250px; text-align:center;" |
'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
1 [[mole]]s of H<sub>2</sub> is needed for every 1 [[mole]] of F<sub>2</sub>
| style="height:20px; width:250px; text-align:center;" |
'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
2 [[mole]]s of H<sub>2</sub> is needed for every 1 [[mole]] of O<sub>2</sub>
| style="height:20px; width:250px; text-align:center;" |
'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
2 [[mole]]s of C<sub>2</sub>H<sub>6</sub> is needed for every 7 [[mole]]s of O<sub>2</sub>
|-
| style="height:20px; width:250px; text-align:center;" |
'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
[[Volume (Space)|Volume]] of H<sub>2</sub> = 24 x (number of moles)
12 = 24 x (number of moles)
Moles of H<sub>2</sub> = 12/24
Moles of H<sub>2</sub> = 0.5 mol
Therefore at a [[ratio]] of 1:1 then 0.5 mol of F<sub>2</sub> are needed.
[[Volume (Space)|Volume]] of F<sub>2</sub> = 24 x (number of moles)
15 = 24 x (number of moles)
Moles of F<sub>2</sub> = 15/24
Moles of F<sub>2</sub> = 0.625 mol
There is more than enough F<sub>2</sub>, therefore H<sub>2</sub> is the '''limiting reactant'''.
| style="height:20px; width:250px; text-align:center;" |
'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
[[Volume (Space)|Volume]] of H<sub>2</sub> = 24 x (number of moles)
18 = 24 x (number of moles)
Moles of H<sub>2</sub> = 18/24
Moles of H<sub>2</sub> = 0.75 mol
Therefore at a [[ratio]] of 2:1 then 0.375 mol of O<sub>2</sub> are needed.
[[Volume (Space)|Volume]] of O<sub>2</sub> = 24 x (number of moles)
8 = 24 x (number of moles)
Moles of O<sub>2</sub> = 8/24
Moles of O<sub>2</sub> = 0.333 mol
There is not enough O<sub>2</sub>, therefore O<sub>2</sub> is the '''limiting reactant'''.
| style="height:20px; width:250px; text-align:center;" |
'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
[[Volume (Space)|Volume]] of H<sub>2</sub> = 24 x (number of moles)
9 = 24 x (number of moles)
Moles of C<sub>2</sub>H<sub>6</sub> = 9/24
Moles of C<sub>2</sub>H<sub>6</sub> = 0.375 mol
Therefore at a [[ratio]] of 2:7 then 1.3125 mol of O<sub>2</sub> are needed.
[[Volume (Space)|Volume]] of O<sub>2</sub> = 24 x (number of moles)
32 = 24 x (number of moles)
Moles of O<sub>2</sub> = 32/24
Moles of O<sub>2</sub> = 1.333 mol
There is more than enough O<sub>2</sub>, therefore C<sub>2</sub>H<sub>6</sub> is the '''limiting reactant'''.
|}
===References===
====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=9e29fad914244909903e5e93f8a01d170 ''Limiting reactants, page 45, GCSE Chemistry; The Revision Guide, 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 ''Limiting reactants, page 67, GCSE Chemistry; Third Edition, Oxford University Press, 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 ''Limiting reactants, pages 114-116, GCSE Combined Science Trilogy; Chemistry, CGP, 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 ''Limiting reactants, pages 124-126, GCSE Chemistry, CGP, AQA '']
:[https://www.amazon.co.uk/gp/product/1471851354/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851354&linkCode=as2&tag=nrjc-21&linkId=9012a0d354024419214fb3ad5ac44ba0 ''Limiting reactants, pages 190-1, GCSE Combined Science Trilogy 1, Hodder, 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 ''Limiting reactants, pages 75-6, GCSE Chemistry, Hodder, AQA '']
====Edexcel====
:[https://www.amazon.co.uk/gp/product/1292120193/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120193&linkCode=as2&tag=nrjc-21&linkId=572df39392fb4200db8391d98ae6314e ''Limiting reactant, page 220, GCSE Combined Science, Pearson Edexcel '']
:[https://www.amazon.co.uk/gp/product/1292120215/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120215&linkCode=as2&tag=nrjc-21&linkId=8f96ddb76196848bafdb124354e4cf77 ''Limiting reactant, page 76, GCSE Chemistry, Pearson, 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 ''Limiting reactants, page 31, GCSE Chemistry; The Revision Guide, CGP, 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 ''Limiting reactants, page 84, GCSE Chemistry, CGP, Edexcel '']
:[https://www.amazon.co.uk/gp/product/1782945741/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945741&linkCode=as2&tag=nrjc-21&linkId=30da4f2178da182547b62a7329d13b57 ''Limiting reactants, page 94, GCSE Combined Science; The Revision Guide, CGP, Edexcel '']
====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 ''Limiting reactants, pages 39, 40, Gateway GCSE Chemistry; The Revision Guide, CGP, 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 ''Limiting reactants, pages 98, 158, Gateway GCSE Chemistry, Oxford, OCR '']