Question

Zinc blende is roasted 8n air. Find :
(i) The number of moles of SO2 liberated by 77 g of ZnS.
(ii) The weight of ZnS required to produce 22.4 litres of SO2 at stp.
[S=32,Zn=65,O=16]
Ans.=(i)0.68 moles (ii)97g
Please solve urgently.

Answer 1

2ZnS + 3O2                   →          2ZnO + 2SO2

molecular mass of Zinc blend (ZnS)=65+32=97g
molecuar mass of So2=32+(16x2)=64g
i) using simple unitary method

       wt.of ZnS                    wt.of So2
         2x97                      2x64

           77g                      2x64
                                     --------- x 77 = 50.80g
                                       2x97
hence there will be 50.80g of So2 on adding 77g ZnS

no. of moles of So2 =given mass of So2            50.80
                                   ----------------------------- =  ---------- = 0.7 ans

                                   molecular mass of So2       64

ii) we know that 1 mole of gas have 22.4L volume irrespective of its molecular mass
hence applying simple unitary method
 as in eq. it is given that for 2 moles of so2(hence its volume is 2x22.4L)
    vol. of So2                       wt. of ZnS
        2x22.4 L                           2x97g

          22.4L                               2x97g
                                                ---------- x 22.4L =97g
                                                  2x22.4L
hence 97g ZnS will be required .

Answer 2

Answer:

bsushsjsijeuehhe e eke en

Explanation:

joqoBVisisvsuveueuvveneidbd