How many grams of I2 are present in a solution if 33.75mL of 0.140M Na2S2O3 solution is needed to titrate the I2solution?
I2 + 2S2O3^2- ==> 2I^- + S4O6^2-
mols S2O3^2- = M x L = ?
1 mol I2 = 2 mol S2O3^2-; therefore, mols S2O3^2- = 2*mols I2.
Then M S2O3^2- = mols S2O3^2-/L S2O3^2-
To determine the number of grams of I2 in the solution, we need to use the stoichiometry of the reaction involving I2 and Na2S2O3.
The balanced equation for the reaction between I2 and Na2S2O3 is:
I2 + 2Na2S2O3 -> 2NaI + Na2S4O6
From the equation, we can see that the ratio between I2 and Na2S2O3 is 1:2.
Given:
Volume of Na2S2O3 solution (V) = 33.75 mL = 0.03375 L
Molarity of Na2S2O3 solution (M) = 0.140 M
Now, we can use the formula:
Moles of Na2S2O3 = Molarity x Volume (in L)
Moles of Na2S2O3 = 0.140 M x 0.03375 L = 0.004725 moles of Na2S2O3
Since the ratio of I2 to Na2S2O3 is 1:2, the moles of I2 present in the solution is half of the moles of Na2S2O3 used.
Moles of I2 = 0.004725 moles / 2 = 0.0023625 moles of I2
The molar mass of iodine (I2) is 253.8 g/mol.
Now, we can use the formula:
Grams of I2 = Moles of I2 x Molar mass of I2
Grams of I2 = 0.0023625 moles x 253.8 g/mol = 0.599 g (rounded to three decimal places)
Therefore, there are approximately 0.599 grams of I2 present in the solution.
To determine the number of grams of I2 present in the solution, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between I2 and Na2S2O3.
The balanced chemical equation for the reaction is:
2Na2S2O3 + I2 → Na2S4O6 + 2NaI
From the equation, we can see that for every 1 mole of I2, we need 2 moles of Na2S2O3. This means that the ratio of moles of I2 to moles of Na2S2O3 is 1:2.
Step 1: Calculate the number of moles of Na2S2O3 used.
Molarity (M) = moles/volume (L)
Given:
Molarity of Na2S2O3 solution (M) = 0.140 M
Volume of Na2S2O3 solution used (L) = 33.75 mL = 0.03375 L
moles of Na2S2O3 = Molarity × Volume
moles of Na2S2O3 = 0.140 M × 0.03375 L
Step 2: Calculate the number of moles of I2 present.
Since the mole ratio between I2 and Na2S2O3 is 1:2, we can write:
moles of I2 = (1/2) × moles of Na2S2O3
Step 3: Convert moles of I2 to grams.
To convert moles of I2 to grams, we need to know the molar mass of I2.
Molar mass of I2 = 2 × (atomic mass of I)
Molar mass of I2 = 2 × 126.9 g/mol
Molar mass of I2 = 253.8 g/mol
grams of I2 = moles of I2 × molar mass of I2
By following these steps, you can calculate the number of grams of I2 present in the solution.