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→Finding a Limiting Reactant
===Finding a Limiting Reactant===
====Limiting Reactants and Mass====
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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]]:
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====Limiting Reactants and Concentration====
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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]]?
There is not enough Mg(OH)<sub>2</sub>, therefore Mg(OH)<sub>2</sub> is the '''limiting reactant'''.
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====Limiting Reactants and Volume====
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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]]?
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'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>H2 + F2->2HF</chem>
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'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>2H2 + O2-> 2H2O</chem>
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'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>2C2H6 + 7O2->4CO2 + 6H2O</chem>
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'''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>
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'''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>
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'''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>
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'''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'''.
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'''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'''.
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'''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'''.
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