MediumChemistryHigh School (11–12)The d- and f-Block Elements › Transition Elements (d-Block)NumericalA-Level IB JEE Main

A solution holds 0.0600 mol of potassium manganate…

Contributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A solution holds $0.0600\ \text{mol}$ of potassium manganate. Carbon dioxide is passed in until all of it has reacted: $3\text{K}_2\text{MnO}_4 + 2\text{CO}_2 \rightarrow 2\text{KMnO}_4 + \text{MnO}_2 + 2\text{K}_2\text{CO}_3$ . Use $158\ \text{g/mol}$ for $\text{KMnO}_4$ and $87\ \text{g/mol}$ for $\text{MnO}_2$ . Find the moles of permanganate formed, its mass, the mass of $\text{MnO}_2$ and the percentage of the manganese that ends as permanganate.

Need a reminder? The key ideaShow

Moles = mass in g divided by molar mass in g/mol, or concentration in mol/L times volume in litres. Take every mole ratio from the balanced equation, and follow the chain one step at a time.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

← All Problems

← A 7.90 g sample of pure potassium permanganate, with a molar mass of 158 g/mol, is heated until it…A 1.50 g sample of sodium nitrite, NaNO₂, is dissolved and made up to 250 mL… →

More onThe d- and f-Block Elements › Transition Elements (d-Block)

Discussion (0)

Loading…