please answer all 3 parts. thank you! This question has maltiple parts Work all

Question

please answer all 3 parts. thank you!

This question has maltiple parts Work all the parts to get the most points a lodine is made by the following reaction 2 NalO,(aq)+5 NaHSO (a) 3 NaHSo4(a)+ +2 Na SO4(aq)+H,00)+ 2(ag) Name the two reactants. NalO, is NaHSO is submi b lodine is made by the following reaction lfyounish to prepare 1.17 kg of12 what masses on al03 and NaHSO: are regsare gNalo Nato NaHSO3 lodine is made by the followng reaction What is the theoretscal yield of l, it you mised1S 0 esf NatOs with 125 mt ofoSS MANaHSO

Explanation / Answer

Ans. #a. NaIO3 is oxidizing agent (getting reduced from oxidation number +5 to 0).

NaHSO3 is reducing agent (getting oxidized from oxidation number +4 to +6)

Note: The terms limiting reactant and reagent in excess can’t be used because the amount of two reactants is not mentioned.

#b. Mass of I2 to be produced = 1.17 kg = 1170.0 g

Moles of I2 to be produced = Mass / Molar mass

                                                = 1170.0 g / (253.80894 g/ mol)

                                                = 4.6097667 mol

# According to the stoichiometry of balanced reaction, 2 mol NaIO3 and 5 mol NaHSO3 are required to produce 1 mol I2.

So,

I. Moles of NaIO3 required = (2 mol NaIO3 / 1 mol I2) x Desired Moles of I2

                                                = (2 mol NaIO3 / 1 mol I2) x 4.6097667 mol

                                                = 9.2195 mol

Now,

            Mass of NaIO3 required = Required moles x Molar mass

                                                = 9.2195 mol x (197.892438 g/ mol)

                                                = 1824.4759 g

                                                = 1.8245 kg

II. Moles of NaHSO3 required = (5 mol NaHSO3 / 1 mol I2) x Desired Moles of I2

                                                = (5 mol NaHSO3 / 1 mol I2) x 4.6097667 mol

                                                = 23.0488 mol

Now,

            Mass of NaHSO3 required = Required moles x Molar mass

                                                = 23.0488 mol x (104.061908 g/ mol)

                                                = 2398.5056 g

                                                = 2.399 kg

#c. In balanced reaction, the theoretical molar ration of reactants is-

            NaIO3 : NaHSO3 = 2 : 5

# Moles of NaIO3 taken = 15.0 g / (197.892438 g/ mol) = 0.075798753 mol

Moles of NaIO3 taken = Molarity x Volume of solution in liters

                                    = 0.853 M x 0.125 L

                                    = 0.106625 mol

Experimental molar ration of reactants is-

NaIO3 : NaHSO3 = 0.075798753 mol : 0.106625 mol = 2 : 2.82

# Comparing the theoretical and experimental molar ratios, the moles of NaHSO3 less than theoretical 5 mol when that of NaIO3 is kept constant at 2 mol.

Therefore, NaHSO3 is the limiting reactant.

# The formation of product follows the stoichiometry of limiting reactant.

So,

Theoretical moles of I2 produced = (2 mol I2 / 5 mol NaHSO3) x Moles of NaHSO3 taken

                                                = (2 mol I2 / 5 mol NaHSO3) x 0.106625 mol

                                                = 0.266563 mol

Now,

Theoretical yield of I2 = Theoretical moles of I2 x Molar mass

                                    = 0.266563 mol x (253.80894 g/ mol)

                                    = 67.656 g

Hence, theoretical yield of I2 = 67.656 g

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