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I would really appreciate some help on this problem. =)

At 218 C, Kc = .00012 for the equilibrium

NH4HS(s)<==> NH3(g)+ H2S(g)

Calculate the equilibrium concentrations of NH3 and H2S if a sample solid NH4HS is placed in a closed vessel and decomposes until equilibrium is reached.

Part (B) If the flask has a volume of 0.500L, what is the minimum mass of NH4HS(s) that must be added to the flask to achieve equilibrium?

Write the Kc expression for the reaction.
Kc = 0.00012=(H2S)(NH3)
Since (H2S)=(NH3), then
(NH3)^2=(H2S)^2 = 0.00012
Solve for (NH3).

For part b, you know concentration of NH3 and H2S in mols/L. In 1/2 L there will be how many mols? And how many mols NH4HS provided this? And what is the mass of NH4HS that provides this many mols? Post your work if you get stuck.

  • AP CHEM -

    So, if you use that equation, you get 0.01095 M for NH3. You multiply that by 0.500L to get the number of moles, and you get 0.005477 mol NH3. Since your ratios are all the same then NH4HS is also 0.005477 mols. Then you multiply that by 51.1g/mol to get the mass of NH4HS. I got 0.28 g of NH4HS but that's not right, so I was wondering if there's anything wrong in my work

  • AP CHEM -

    You don't have enough SWAG in yout equations. You have to use the formula swag/yolo * eqm^SWAG = YOU to get the answer

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