An alternative method for producing hydriodic acid is the reaction of iodine with hydrogen sulfide:
H2S + I2 ----> 2 HI +S
a) How many grams of I2 are needed to react with 49.2g of H2S?
b) How many grams of HI are produced from the reaction of 95.4g of H2S with excess I2?
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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 per mol
= 367.538 g.
95.2 g of H2S reacts with excess I2. This means all the H2S is converted into HI and S.
Therefore number of moles in 95.2 g of H2S = 95.2 / 34 mol
= 2.8 mol
According to stoichiometry we get 2 mols of HI per one mol of H2S reacted
Therefore HI mol generated = 2 x 2.8 mol = 5.6 mol
So the mass of HI produced = 5.6 mol x 128 g per mol
= 716.8 g.
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