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`

For the reaction : SO_(2)+2H_(2)Sto3S+2H_(2)O -

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:

Calculate the oridation numbers of sulphur in H_(2)SO_(5) and in H_(2)S_(2)O_(8)

H_(2)S reacts with SO_(2) to form

Calculate the oxidation number of sulphur in H_(2)SO_(3) and in H_(2)S_(2)O_(8)

Equivalent mass of oxidizing agent in the reaction, SO_(2) +2H_(2)S = 3S + 2H_(2)O is

Equal volumes of H_(2)S & SO_(2) react at NTP to form H_(2)O and S. 2H_(2)S+SO_(2) to 2H_(2)O+3S . In this reaction, I. H_(2)S is the limiting reactant. II. SO_(2) is the limiting reactant. III. Sulphur formed is three times of SO_(2) reacted IV. sulphur formed is 1.5 times of H_(2)S reacted. Select the correct statements(s).

H_(2)S reacts with SO_(3) to form