How many grams of I2 are present in a solution if 39.85mL of 0.160M Na2S2O3 solution is needed to titrate the I2 Solution
To find the number of grams of I2 in the solution, you will use the given information of the volume and molarity of Na2S2O3 solution used for titration.
First, you need to understand the balanced chemical equation between Na2S2O3 and I2:
2Na2S2O3 + I2 -> Na2S4O6 + 2NaI
From the balanced equation, you can see that two moles of Na2S2O3 react with one mole of I2.
Now, use the molarity and volume of Na2S2O3 to calculate the number of moles of Na2S2O3 used:
Moles of Na2S2O3 = Volume of Na2S2O3 solution (in liters) × Molarity of Na2S2O3
Convert the volume of Na2S2O3 solution given from milliliters to liters:
Volume of Na2S2O3 solution = 39.85 mL ÷ 1000 mL/L = 0.03985 L
Now, substitute the values into the formula:
Moles of Na2S2O3 = 0.03985 L × 0.160 M = 0.006376 moles
Since the stoichiometry of the reaction indicates that two moles of Na2S2O3 react with one mole of I2, half the number of moles of Na2S2O3 is equal to the number of moles of I2:
Moles of I2 = 0.006376 moles ÷ 2 = 0.003188 moles
Finally, to find the mass of I2, you need to use the molar mass of I2, which is 253.8 g/mol.
Grams of I2 = Moles of I2 × Molar mass of I2
Grams of I2 = 0.003188 moles × 253.8 g/mol ≈ 0.809 grams
Therefore, there are approximately 0.809 grams of I2 present in the solution.