Question

A beaker containing 50 cm³ of 0.2 mole/dm³ sodium thiosulfate was placed on the top of a black cross. Then, 5 cm³ of 2 moles/dm³ hydrochloric acid was added. And the clock started. Which (Na2S2O3 or HCl) is the limiting reagent?

A) Na2S2O3
B) HCl
C) Both Na2S2O3 and HCl
D) None of them

Answer 1

The limiting reactant in the reaction between sodium thiosulfate and hydrochloric acid, given the provided volumes and concentrations, is hydrochloric acid (HCl). This is because there are fewer moles of HCl available than are necessary to react with the available moles of sodium thiosulfate based on the stoichiometry of the reaction.

Step-by-step explanation:

To determine which reagent is limiting, we need to calculate the number of moles of each reactant and compare them based on the stoichiometry of the reaction.

For sodium thiosulfate (Na2S2O3):

• Volume = 50 cm³ or 0.05 dm³
• Concentration = 0.2 mol/dm³
• Moles of Na2S2O3 = Volume × Concentration = 0.05 dm³ × 0.2 mol/dm³ = 0.01 moles

For hydrochloric acid (HCl):

• Volume = 5 cm³ or 0.005 dm³
• Concentration = 2 mol/dm³
• Moles of HCl = Volume × Concentration = 0.005 dm³ × 2 mol/dm³ = 0.01 moles

The reaction between sodium thiosulfate and hydrochloric acid proceeds as:

Na2S2O3 + 2 HCl → 2 NaCl + SO2 + S + H2O

According to the stoichiometry of the reaction, 1 mole of Na2S2O3 reacts with 2 moles of HCl. Since we have equal moles of both reactants, and the reaction requires twice as many moles of HCl per mole of Na2S2O3, HCl is the limiting reactant.