Question

A 3.41 g sample of H₂S and a 3.33 g sample of SO₂ react in a closed 4.31 L container at 479 K, according to the following balanced chemical equation:

16H₂S(g) + 8SO₂(g) → 3S₈(g) + 16H₂O(l)

Calculate the P_S8 (in atm) in the container after the reaction has gone to completion.

Answer 1

Solution Given data: mass of H₂S = 3.41g mass of so₂ = 3.33g Volume of container (v) = 4.31 L Temperature in container (T)=473 R o . molar mass of H₂S = 34.1 a/mol mai So. mol of HS > maxs_ 3.419 = 0.1 m ) moluk mass/H,S 34.13/mol Similarly so molar mass of 50₂ = 64.066 g/mol mol of 502 = ( mure mass)so, 64.066.3fmos 3.338_ =0-052 mol Given Balance reaction 16H2 519)+ 8502(9)→ 352(9+164,011) from above eqn 8 mol so need 16 mol H₂S . 1 mol soz need 2 mol H₂S Here 0.052 mol SO₂ need 2x0.052 mol H₂S => 0.104 mol Has But H₂S is only o-l mol ,so H₂S is limiting reagent s Reaction will complete till H₂s read completely Here from reaction - 16 mol H₂S produce = 3 mol Sg cm 1 mol H2S produce = pamol 58 $0 0.1 mol H2S produce = Xol mol S8 20-01875 mal 58 so mol of 58 produced (n) = 0.01875 mol from Ideal gas equation . PV = nRT ar Put all values de R=0.0821 homestamp 10.01875 Tot (0.0821 L-am ) (4796) Tolk → Sg 4.31% Psa = 0.171 odm Answer