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
Answer: The mass of iodine gas needed to react is 477.14 g
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of [tex]N_2H_4[/tex] = 30.1 g
Molar mass of [tex]N_2H_4[/tex] = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }N_2H_4=\frac{30.1g}{32g/mol}=0.94mol[/tex]
The chemical equation for the reaction of [tex]N_2H_4[/tex] and iodine gas follows:
[tex]2I_2+N_2H_4\rightarrow 4HI+N_2[/tex]
By Stoichiometry of the reaction:
1 mole of [tex]N_2H_4[/tex] reacts with 2 moles of iodine gas
So, 0.94 moles of [tex]N_2H_4[/tex] will react with = [tex]\frac{2}{1}\times 0.94=1.88mol[/tex] of iodine gas
Now, calculating the mass of iodine gas by using equation 1, we get:
Molar mass of iodine gas = 253.80 g/mol
Moles of iodine gas = 1.88 moles
Putting values in equation 1, we get:
[tex]1.88mol=\frac{\text{Mass of iodine gas}}{253.80g/mol}\\\\\text{Mass of iodine gas}=(1.88mol\times 253.80g/mol)=477.14g[/tex]
Hence, the mass of iodine gas needed to react is 477.14 g
