Given :
Initial concentration of nitric acid, 
.
Final concentration of nitric acid, 
.
Volume of 
, 
To Find :
Volume needed( 
).
Solution :
We know,
Hence, this is the required solution.
5.0x10^1 kg is the correct answer
The reaction is,
H2S + I2 --------------> 2 HI +S
Molar weight of H2S = 34 g per mol
Molar weight of HI =128 g per mol
Molar weight of I2 =254 g per mol
Moles of H2S in 49.2 g = 49.2 /34 mol = 1.447 mol
So according to stoichiometry of the reaction, number of I2 mols needed
= 1.447 mol
The mass of I2 needed = 1.447 mol x 254 g
