Question #9c730

1 Answer
May 22, 2017

Answer:

There are 0.0248 mol of #"H"_2"S"# at equilibrium

Explanation:

The equation for the equilibrium is

#"NH"_4"HS(s)" ⇌ "NH"_3"(g)" + "H"_2"S(g)"#

#K = ["NH"_3]["H"_2"S"] = 1.80 × 10^"-4"#

Calculate the equilibrium concentration of #"NH"_3"#

#["NH"_3] = "0.279 mol"/"6.20 L" = "0.0450 mol/L"#

Calculate the equilibrium concentration of #"H"_2"S"#

#0.0450 × ["H"_2"S"] = 1.80 × 10^"-4"#

#["H"_2"S"] = (1.80 × 10^"-4"}/0.0450 = "0.004 00 mol/L"#

Calculate the moles of #"H"_2"S"#

#"Moles H"_2"S" = 6.20 color(red)(cancel(color(black)("L"))) × "0.004 00 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.0248 mol"#