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→Finding a Limiting Reactant
[[Glucose]] is the [[Limiting Reactant|limiting reactant]].
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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]]?
| 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]]?
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'''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>
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'''1. Write the [[Balanced Symbol Equation]] for the [[Chemical Reaction|reaction]].'''
<chem>2HCl+Mg(OH)2->MgCl2+2H2O</chem>
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'''2. State the [[ratio]] of [[mole]]s of each [[chemical]] needed.'''
1 [[mole]]s of HCl is needed for every 1 [[mole]] of NaOH
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'''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
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'''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>
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'''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'''.
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'''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'''.
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'''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'''.
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