Question

Saturated solutions are often prepared by bubbling a gas through water. At 298 K and 1 atm, SO2 is bubbled through H2O to form a saturated solution. A 10 mL sample of the solution is reacted with potassium iodate and the I2 formed is then reacted completely with 65.6 mL of a 0.10 molar solution of Na2SO3, Calculate moles of iodine produced.

Answer 1

To calculate the moles of iodine, use the volume and molarity of the Na2SO3 solution. Given that 65.6 mL of a 0.10 M Na2SO3 solution reacted completely, the moles of iodine produced are 0.00656 mol, as the reaction with iodine is 1:1.

Step-by-step explanation:

The question is asking to calculate the moles of iodine produced in a reaction sequence involving potassium iodate and a solution of sodium sulfite. To find the moles of iodine produced, we can use the volume and molarity of the Na2SO3 solution that reacts completely with the iodine.

Since we know that the Na2SO3 solution is 0.10 M and 65.6 mL of it is used, we can first find the number of moles of sodium sulfite that have reacted:

Moles of Na2SO3 = 0.10 mol/L × 0.0656 L = 0.00656 mol

Now, knowing the stoichiometry of the reaction, where sodium sulfite reacts with iodine in a 1:1 molar ratio, the moles of iodine will be the same as the moles of sodium sulfite.

Therefore, the moles of iodine produced are also 0.00656 mol.