Limiting Reactant - Revision history

Ellen References at 08:53, 14 December 2019

2019-12-14T08:53:43Z

Ellen References at 14:47, 22 November 2019

2019-11-22T14:47:33Z

Ellen References at 09:29, 8 November 2019

2019-11-08T09:29:05Z

NRJC: /* Limiting Reactants and Mass */

2019-01-23T15:59:49Z

‎Limiting Reactants and Mass

NRJC: /* Limiting Reactants and Concentration */

2019-01-23T15:57:46Z

‎Limiting Reactants and Concentration

NRJC: /* Finding a Limiting Reactant */

2019-01-23T15:56:37Z

‎Finding a Limiting Reactant

NRJC: /* Finding a Limiting Reactant */

2019-01-23T15:01:20Z

‎Finding a Limiting Reactant

NRJC: /* Finding a Limiting Reactant */

2019-01-23T15:00:13Z

‎Finding a Limiting Reactant

NRJC: /* Finding a Limiting Reactant */

2019-01-23T14:33:36Z

‎Finding a Limiting Reactant

NRJC: /* Finding a Limiting Reactant */

2019-01-23T14:32:18Z

‎Finding a Limiting Reactant

← Older revision		Revision as of 08:53, 14 December 2019
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	:[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/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 '']		:[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 '']

← Older revision		Revision as of 14:47, 22 November 2019
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	:[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/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 '']		:[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 '']

← Older revision		Revision as of 09:29, 8 November 2019
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	There is more than enough O<sub>2</sub>, therefore C<sub>2</sub>H<sub>6</sub> is the '''limiting reactant'''.		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 '']

@@ Line 76: / Line 76: @@
 No. [[Mole]]s = 0.5 mol
-Therefore 0.5 [[mole]]s of O<sub>2</sub> is needed.
+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>
@@ Line 95: / Line 95: @@
 No. [[Mole]]s = 1 mol
-Therefore 0.75 [[mole]]s of O<sub>2</sub> are needed.
+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>
@@ Line 114: / Line 114: @@
 No. [[Mole]]s = 1 mol
-Therefore 6 [[mole]]s of O<sub>2</sub> are needed.
+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>

@@ Line 173: / Line 173: @@
 Moles of HCl = 0.05 mol
-Therefore 0.05 mol of NaOH are needed.
+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>
@@ Line 191: / Line 191: @@
 Moles of H<sub>2</sub>SO<sub>4</sub> = 0.06 mol
-Therefore 0.12 mol of NaOH are needed.
+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>
@@ Line 209: / Line 209: @@
 Moles of HCl</sub> = 0.15 mol
-Therefore 0.075 mol of Mg(OH)<sub>2</sub> are needed.
+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>

@@ Line 14: / Line 14: @@
 ===Finding a Limiting Reactant===
 {| class="wikitable"
 |-
 | 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]]:
@@ Line 126: / Line 126: @@
 |}
 {| class="wikitable"
 |-
 | 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]]?
@@ Line 218: / Line 218: @@
 There is not enough Mg(OH)<sub>2</sub>, therefore Mg(OH)<sub>2</sub> is the '''limiting reactant'''.
 |}

@@ Line 138: / Line 138: @@
 | 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>
+<chem>HCl + NaOH->NaCl + H2O</chem>
 | style="height:20px; width:250px; text-align:center;" |
 '''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
-<chem>H2SO4+2NaOH->Na2SO4+2H2O</chem>
+<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>
+<chem>2HCl + Mg(OH)2->MgCl2 + 2H2O</chem>
 |-
 | style="height:20px; width:250px; text-align:center;" |

← Older revision		Revision as of 15:00, 23 January 2019
Line 124:		Line 124:
	[[Glucose]] is the [[Limiting Reactant\|limiting reactant]].		[[Glucose]] is the [[Limiting Reactant\|limiting reactant]].

		+	\|}
		+
		+
		+	{\| class="wikitable"
		+	\|+Limiting Reactants and Concentration
		+	\|-
		+	\| 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;" \|If 0.12 dm<sup>3</sup> of 0.5[[Molarity\|M]] [[Sulphuric Acid]] is added to 0.1 dm<sup>3</sup> of 1.1[[Molarity\|M]] [[Sodium Hydroxide]] which is the [[Limiting Reactant\|limiting reactant]] in this [[Chemical Reaction\|reaction]]?
		+
		+	\| style="height:20px; width:250px; text-align:center;" \|If 0.1 dm<sup>3</sup> of 1.5[[Molarity\|M]] [[Hydrochloric Acid]] is added to 0.300 dm<sup>3</sup> of 0.2[[Molarity\|M]] [[Magnesium 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>
		+
		+	\| style="height:20px; width:250px; text-align:center;" \|
		+	'''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;" \|
		+	'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
		+
		+	1 [[mole]]s of HCl is needed for every 1 [[mole]] of NaOH
		+
		+	\| 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>SO<sub>4</sub> are needed for every 2 [[mole]]s of NaOH
		+
		+	\| style="height:20px; width:250px; text-align:center;" \|
		+	'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
		+
		+	2 [[mole]]s of HCl is needed for every 1 [[mole]] of Mg(OH)<sub>2</sub>
		+	\|-
		+
		+	\| style="height:20px; width:250px; text-align:center;" \|
		+	'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
		+
		+	'' '''concentration of HCl''' '' = <math>\frac{Moles (mol)}{volume (dm^3)}</math>
		+
		+	0.5 = <math>\frac{Moles}{0.1}</math>
		+
		+	Moles of HCl = 0.05 mol
		+
		+	Therefore 0.05 mol of NaOH are needed.
		+
		+	'' '''concentration of NaOH''' '' = <math>\frac{Moles (mol)}{volume (dm^3)}</math>
		+
		+	0.6 = <math>\frac{Moles}{0.1}</math>
		+
		+	Moles of NaOH = 0.06 mol
		+
		+	There is more than enough NaOH, therefore HCl is the '''limiting reactant'''.
		+	\| style="height:20px; width:250px; text-align:center;" \|
		+	'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
		+
		+	'' '''concentration of H<sub>2</sub>SO<sub>4</sub>''' '' = <math>\frac{Moles (mol)}{volume (dm^3)}</math>
		+
		+	0.5 = <math>\frac{Moles}{0.12}</math>
		+
		+	Moles of H<sub>2</sub>SO<sub>4</sub> = 0.06 mol
		+
		+	Therefore 0.12 mol of NaOH are needed.
		+
		+	'' '''concentration of NaOH''' '' = <math>\frac{Moles (mol)}{volume (dm^3)}</math>
		+
		+	1.1 = <math>\frac{Moles}{0.1}</math>
		+
		+	Moles of NaOH = 0.11 mol
		+
		+	There is not enough NaOH, therefore NaOH is the '''limiting reactant'''.
		+	\| style="height:20px; width:250px; text-align:center;" \|
		+	'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
		+
		+	'' '''concentration of HCl''' '' = <math>\frac{Moles (mol)}{volume (dm^3)}</math>
		+
		+	1.5 = <math>\frac{Moles}{0.1}</math>
		+
		+	Moles of HCl</sub> = 0.15 mol
		+
		+	Therefore 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>
		+
		+	0.2 = <math>\frac{Moles}{0.3}</math>
		+
		+	Moles of Mg(OH)<sub>2</sub> = 0.06 mol
		+
		+	There is not enough Mg(OH)<sub>2</sub>, therefore Mg(OH)<sub>2</sub> is the '''limiting reactant'''.
	\|}		\|}

@@ Line 31: / Line 31: @@
 |-
 | style="height:20px; width:250px; text-align:center;" |
-Find the [[Relative Formula Mass]] of the [[reactant]]s.
+'''1. Find the [[Relative Formula Mass]] of the [[reactant]]s.'''
 M<sub>r</sub> of C = 12g
@@ Line 38: / Line 38: @@
 | style="height:20px; width:250px; text-align:center;" |
-Find the [[Relative Formula Mass]] of the [[reactant]]s.
+'''1. Find the [[Relative Formula Mass]] of the [[reactant]]s.'''
 M<sub>r</sub> of NaOH = 40g
@@ Line 45: / Line 45: @@
 | style="height:20px; width:250px; text-align:center;" |
-Find the [[Relative Formula Mass]] of the [[reactant]]s.
+'''1. Find the [[Relative Formula Mass]] of the [[reactant]]s.'''
 M<sub>r</sub> of NaOH = 40g
@@ Line 52: / Line 52: @@
 |-
 | style="height:20px; width:250px; text-align:center;" |
-State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
+'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
 [[mole]]s of C is needed for every 1 [[mole]] of O<sub>2</sub>
 | style="height:20px; width:250px; text-align:center;" |
-State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
+'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
 [[mole]]s of Al are needed for every 3 [[mole]]s of O<sub>2</sub>
 | style="height:20px; width:250px; text-align:center;" |
-State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
+'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
 [[mole]] of C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> is needed for every 6 [[mole]]s of O<sub>2</sub>
@@ Line 68: / Line 68: @@
 | style="height:20px; width:250px; text-align:center;" |
-Find the number of [[mole]]s supplied of each [[chemical]].
+'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
 No. [[Mole]]s of C = <math>\frac{Mass}{M_r}</math>
@@ Line 87: / Line 87: @@
 [[mole]] of O<sub>2</sub> = 32g
 | style="height:20px; width:250px; text-align:center;" |
-Find the number of [[mole]]s supplied of each [[chemical]].
+'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
 No. [[Mole]]s of Al = <math>\frac{Mass}{M_r}</math>
@@ Line 106: / Line 106: @@
 | style="height:20px; width:250px; text-align:center;" |
-Find the number of [[mole]]s supplied of each [[chemical]].
+'''3. Find the number of [[mole]]s supplied of each [[chemical]].'''
 No. [[Mole]]s of C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> = <math>\frac{Mass}{M_r}</math>