Question #9c730

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Answer 1 · 2017-05-22T18:34:23

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

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 $"H"_2"S"$

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