Question #508e4

Sodium thiosulfate, #"Na"_2"S"_2"O"_3#, will react with hydrochloric acid, #"HCl"#, to form solid sulfur, #"S"#, gaseous sulfur dioxide, #"SO"_2#, aqueous sodium chloride, #"NaCl"#, and water.

The key here is the balanced chemical equation that describes this redox reaction

#"Na"_ 2"S"_ 2"O"_ (3(aq)) + color(red)(2)"HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "S"_ ((s)) darr + "SO"_ (2(g)) uarr + "H"_ 2"O"_((l))#

Notice that the reaction consumes #color(red)(2)# moles of hydrochloric acid for every #1# mole of sodium thiosulfate.

This means that in order for the reaction to consume both reactants completely, you must have twice as many moles of hydrochloric acid than you have of sodium thiosulfate.

Now, you're mixing equal volumes, #"50 cm"^(3)#, of two solutions of equal molarities, #"0.1 mol dm"^(-3)#, which of course implies that you're dealing with equal numbers of moles of sodium thiosulfate and hydrochloric acid.

This means that you have fewer moles of hydrochloric acid than you'd need to allow for all the moles of sodium thiosulfate to react #-># hydrochloric acid will act as a limiting reagent.

The reaction will stop because the hydrochloric acid will be completely consumed before all the moles of sodium thiosulfate get the chance to take part in the reaction.

As you can see in the video below, this reaction is often used to show how concentration affects reaction rate