EasyChemistryHigh School (11–12)The d- and f-Block ElementsNumerical

A 25.0 mL sample of 0.0800 mol/L sodium chromate solution is acidified with 0.100 mol/L sulfuric…

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A $25.0\ \text{mL}$ sample of $0.0800\ \text{mol/L}$ sodium chromate solution is acidified with $0.100\ \text{mol/L}$ sulfuric acid until all the chromate has become dichromate: $2\text{Na}_2\text{CrO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$ . Find the amount of chromate, the amount of dichromate formed and the volume of sulfuric acid needed.

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

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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.

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