Answer:
There were 0.268 grams I2 present
Explanation:
Step 1: Data given
Volume of Na2SO3 = 42.25 mL = 0.04225 L
Molarity of Na2SO3 = 0.0500 M
Volume of iodine = 150.0 mL
Step 2: The balanced equation
I2 + 2 Na2SO3 → Na2S2O6 + 2 NaI
Step 3: Calculate moles of Na2SO3
Moles Na2SO3 = molarity * volume
Moles Na2SO3 = 0.0500 M * 0.04225 L
Moles Na2SO3 = 0.0021125 moles
Step 4: Calculate moles I2
For 1 mol I2 we need 2 moles Na2SO3
For 0.0021125 moles Na2SO3 we have 0.00105625
Step 5: Calculate mass of I2
Mass I2 = moles * molar mass
Mass I2 = 0.00105625 * 253.8 g/mol
Mass I2 = 0.268 grams
There were 0.268 grams I2 present
