Question

Question #9c730

Answer 1

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"`