Stoichiometry of Redox and Acid-Base Reactions
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2. Acidified potassium Manganate (VII) solution oxidises Iron (II) to Iron (III) as shown in the ionic equation below. If 0.2M $KMnO_4$ solution is needed to react with $25cm^3$ of 0.1M Iron (II) ammonium Sulphate, Calculate the volume of $KMnO_4$ solution required. (3mks) $$MnO^-_{4(aq)} + 5Fe^{2+}_{(aq)} + 8H^+_{(aq)} \rightarrow Mn^{2+}_{(aq)} + 4H_2O_{(l)} + 5Fe^{3+}_{(aq)}$$ (b) 12g of a mixture of sodium Sulphate and Sodium Carbonate were mixed with distilled water in a flask and topped up to $100cm^3$. $25cm^3$ of this solution required $12.5cm^3$ of 0.2M Sulphuric (VI) acid for complete reaction. (i) Write down the chemical equation for the reaction that occurred between the mixture and Sulphuric (VI) acid . (1mk) (ii) Calculate the number of moles of $H_2SO_4$ which reacted with the mixture. (2mks) (iii) Determine the number of moles of the substances in the mixture that reacted with $H_2SO_4$ (2mks)
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Hi Jackline, let's work through this chemistry stoichiometry problem together. We'll start with part a, which involves a redox reaction between manganate seven ions and iron two ions.
Part (a): Redox Titration
The balanced ionic equation shows the mole ratio between the reactants. Let's write it down and look at the stoichiometry.
From the equation, notice that one mole of manganate reacts with five moles of iron two. This one to five ratio is crucial for our calculation.
First, we calculate the number of moles of iron two ammonium sulphate using its volume and molarity. We have twenty-five cubic centimeters of point one molar solution.
We convert volume to liters by dividing by one thousand. So, it is point one times twenty-five over one thousand.
This gives us zero point zero zero two five moles of iron two ions.
Now, using the one to five mole ratio, the moles of potassium manganate required will be one-fifth of the moles of iron.
Finally, we calculate the volume of zero point two molar KMnO4 needed. Volume equals moles divided by molarity.
Plugging in our values, we get zero point zero zero zero five divided by zero point two.
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