A 30 L of air sample (density of 1.2 g/L) was passed through an absorption tower

Question

A 30 L of air sample (density of 1.2 g/L) was passed through an absorption tower containing a solution of Cd 2+ where H2S was retained as CDs. The mixture was acidified and treated with 10.0 mL of 0.01070 M I2. After the reaction S2- + I2 = S(s) + 2I-. Calculate the H2S in ppm. A 30 L of air sample (density of 1.2 g/L) was passed through an absorption tower containing a solution of Cd 2+ where H2S was retained as CDs. The mixture was acidified and treated with 10.0 mL of 0.01070 M I2. After the reaction S2- + I2 = S(s) + 2I-. Calculate the H2S in ppm.

Explanation / Answer

moles of I2 in 10ml of 0.01070M= 0.01070*10/1000 =0.000107

moles of S2- = moles of I2= 0.000107

H2S------>S2- + 2H+

Moles of H2S= 0.000107

mass of H2S= moles* molar mass = 0.000107*34 =0.003638 gm

mass of air = Volume*density = 30L*1.2 g/L= 36 gm

ppm of H2S= 0.003638/36 =1.01*10-4 gm H2S/gm= 101*10-6 gm/gm air = 101 ppm

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