Question

12 Sulfuric acid is a very important is a good of its induni ommodity chemical., anod indeed, a nationfr fertl or an array of applications including: detergents. Most of the Sulfuric scid is process used to make sulfuric acid is shown below: (12 th Sulfuric scid is used for fe treatment Step 1 Step 2 1r 42.59 g of Hs . How many moles of which reactant were in excess? was reacted with 56.0 g of O: (7 pts.) O, and H:O In step 2 of the reaction, you are able to recover 62.8 grams of SO, and react that with exces to form 83.4 grams of HSO (5 pts.) b. What is the percent yield of step 27 What is the overall percent yield of the reaction? c.

Answer 1

The reactions given are -

2H2S + 3O2 -> 2H2O + 2SO2 -----Step1

2SO2 + 2H2O + O2 -> 2H2SO4 -----Step2

To get the net overall reaction we must add both of these steps to get

2H2S + 3O2 + 2SO2 + 2H2O + O2 -> 2H2O + 2SO2 + 2H2SO4

Cancelling out common terms of both the sides we get -

2H2S + 4O2 -> 2H2SO4

Cancelling the coefficient 2 from both the sides we get final equation as -

H2S + 2O2 -> H2SO4

So, eveymole of H2S needs 2 moles of O2 to react completely and vice-versa.

We also have 42.59 g of H2S and molar mass of H2S = molar mass of H*2 + molar mass of S = 1*2 + 32*1 = 34 g/mol so, moles of H2S in 42.59 g = weight of H2S/molar mass of H2S = 42.59/34 = 1.252 moles

We also have 56 g of O2 and molar mass of O2 = molar mass of O*2 = 16*2 = 32 g/mol so, moles of O2 in 56 g = weight of O2/molar mass of O2 = 56/32 = 1.75 moles

But we needed 2*1.252 moles of O2 for complete reaction as eveymole of H2S needs 2 moles of O2 to react completely.

So, number of moles of O2 are limited in number or H2S is excess in number by 1.252 - 1.75/2 = 0.377 moles.

b)

In step 2 we have 2SO2 + 2H2O + O2 -> 2H2SO4

no. of moles in 62.8 g of SO2 = weight of SO2/molar mass of SO2 , molar mass of SO2 = molar mass of S + 2*molar mass of O = 32 + 2*16 = 64 g/mol so, moles of SO2 = 62.8/64 = 0.98125 moles

Excess of O2 and H2O is taken so, theoretically. every mole of SO2 must produce 1 mole of H2SO4 (ratio of SO2 and H2SO4 in the balanced chemical equation is 1/1)

So, 0.98125 moles of H2SO4 must be produced but we have 83.4 g of H2SO4, moles in 83.4 g of H2SO4 = weight of H2SO4/molar mass of H2SO4, molar mass = 2*molar mass of H + molar mass of S + molar mass of O*4 = 2*1 + 32 + 16*4 = 98 g/mol so, 83.4 g of H2SO4 = 83.4/98 = 0.851 moles.

So, percentage yield = (experimental yield/theoretical yield)*100 = (0.851/0.98125)*100 = 86.728 %

For overall reaction -

H2S + 2O2 -> H2SO4

Every mole of H2SO4 is made by 2 moles of O2 and moles of O2 taken are 1.75 moles so, the moles of H2SO4 produced must be 1.75/2 = 0.875 moles theoretically.

But in experiment only 0.851 moles of H2SO4 are produced so,

So, percentage yield = (experimental yield/theoretical yield)*100 = (0.851/0.875)*100 = 97.257 %