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2 mol of H(2)S and 11.2 L of SO(2) at N....

2 mol of `H_(2)S` and 11.2 L of `SO_(2)` at N.T.P. react to form x moles of sulphur, x is
`SO_(2)+2H_(2)S to 3S +2H_(2)O`

A

1.5

B

3

C

11.2

D

6

Text Solution

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The correct Answer is:
a
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SO_(2)+H_(2)S to

1 mol of SO_(2) and 1 mol of H_(2)S react completely to form H_(2)O and S as follows: SO_(2)+2H_(2)S to 2H_(2)O+3S (At. mass S = 32, O = 16) The mass of S obtained is:

underline(S)O_(2)+H_(2)O to H_(2)SO_(3)

In the reaction, H_(2)S+H_(2)SO_(4)toS+2H_(2)O

underline(S)O_(2)Cl_(2)+H_(2)O to H_(2)SO_(4)+HCl

underline(S)Ocl_(2)+H_(2)O to H_(2)SO_(3)+HCl

Under S.T.P. 1 mol of N_(2) and 3 mol of H_(2) will form on complete reaction