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lianna [129]
3 years ago
5

g For the reaction Ag2S(s)⇌2Ag+(aq)+S2−(aq)Ag2S(s)⇌2Ag+(aq)+S2−(aq), Keq=2.4×10−4Keq=2.4×10−4, and the equilibrium concentration

of sulfide ion is [S2−]=0.0023M[S2−]=0.0023M. What is [Ag+]atequilibrium?
Chemistry
1 answer:
Roman55 [17]3 years ago
8 0

Answer:

0.32 M

Explanation:

Step 1: Write the balanced reaction at equilibrium

Ag₂S(s) ⇌ 2 Ag⁺(aq) + S²⁻(aq)

Step 2: Calculate the concentration of Ag⁺ at equilibrium

We will use the formula for the concentration equilibrium constant (Keq), which is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

Keq = [Ag⁺]² × [S²⁻]

[Ag⁺] = √{Keq / [S²⁻]}

[Ag⁺] = √{2.4 × 10⁻⁴ / 0.0023} = 0.32 M

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Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with
emmasim [6.3K]

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

  • It is a stichiometry problem.

<em>2Na + 2H₂O → 2NaOH + H₂,</em>

  • The balanced equation shows that <em>2.0 moles of Na metal </em>react with 2.0 moles of water to produce 2.0 moles of NaOH and <em>1.0 mole of H₂</em>,
  • Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: <em>PV = nRT</em>,

where, P is the pressure of the gas in atm<em> (P at STP = 1.0 atm)</em>,

V is the volume of the gas in L <em>(V = 7.80 L)</em>,

n is the number of moles in mole,

R is the general gas constant<em> (R = 0.082 L.atm/mol)</em>,

T is the temperature of the gas in K <em>(T at STP = 0.0 °C + 273 = 273.0 K)</em>.

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

<em>Using cross multiplication:</em>

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.

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3 years ago
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