Question

Based on the balanced chemical equation shown below, what volume of 0.250 M K2S₂O₃(aq) is needed to completely react with 24.88 mL of 0.125 M KI3(aq), according to the chemical equation: 2 S₂O₃²-(aq) + 3 i-(aq) → S4O6²-(aq) + 3 i-(aq)?

1) 24.9 mL
2) 12.4 mL
3) 99.5 mL
4) 6.22 mL

Answer 1

To react completely with 24.88 mL of 0.125 M KI3(aq), 24.88 mL of 0.250 M K2S2O3(aq) is needed, as calculated using stoichiometry and the molar ratios from the given balanced chemical equation.

Step-by-step explanation:

To find the volume of 0.250 M K2S2O3(aq) needed to completely react with 24.88 mL of 0.125 M KI3(aq) based on the chemical equation 2 S2O32-(aq) + I2(aq) → S4O62-(aq) + 2 I-(aq), we need to first determine the number of moles of KI3 and then use stoichiometry to find the required moles of K2S2O3 using the molar ratios provided in the equation.

First, calculate the moles of KI3:

moles KI3 = volume KI3 x molarity KI3 = 0.02488 L x 0.125 mol/L = 0.00311 mol I2

Next, use the stoichiometric ratios from the equation to find moles of K2S2O3 needed:

moles K2S2O3 = (0.00311 mol I2) x (2 mol S2O32- / 1 mol I2) = 0.00622 mol S2O32-

Lastly, calculate the volume of K2S2O3:

volume K2S2O3 = moles K2S2O3 / molarity K2S2O3 = 0.00622 mol S2O32- / 0.250 mol/L = 0.02488 L or 24.88 mL

Therefore, the volume of 0.250 M K2S2O3 required is 24.88 mL.