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elena55 [62]
3 years ago
10

A sample of solid NH 4HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (at

m) of ammonia, assuming that some solid NH 4HS remains.
Chemistry
1 answer:
UNO [17]3 years ago
5 0

Answer:

The answer to the question is

The equilibrium partial pressure (atm) of ammonia, assuming that some solid NH₄HS remains 0.26 atm.

Explanation:

To solve the question, we write out the chemical equation as follows

NH₄HS (s) ⇄ NH₃ (g) + H₂S (g)

From the above equation, it is observed that only the gaseous products contribute to the partial pressure

Kp =PNH₃·PH₂S where at Kp = 0.070 and PNH₃, PH₂S are the partial pressures of the gases

However since the number of moles of both gases are equal, therefore by Avogadro's law PNH₃ = PH₂S

Then PNH₃  = √(0.07) = PH₂S = 0.2645 atm. ≅ 0.26 atm.

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New questionWhat mass of calcium chloride (CaCl₂) would beproduced from the reaction of 125.9 g of hydrochloriacid (HCI) with ex
marshall27 [118]

Answer:

191.6 g of CaCl₂.

Explanation:

What is given?

Mass of HCl = 125.9 g.

Molar mass of CaCl₂ = 110.8 g/mol.

Molar mass of HCl = 36.4 g/mol.

Step-by-step solution:

First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O.

Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:

125.9\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=3.459\text{ moles HCl.}

Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is:

3.459\text{ moles HCl}\cdot\frac{1\text{ mol CaCl}_2}{2\text{ moles HCl}}=1.729\text{ moles CaCl}_2.

The final step is to find the mass of CaCl₂ using the molar mass of CaCl₂. This conversion will look like this:

1.729\text{ moles CaCl}_2\cdot\frac{110.8\text{ g CaCl}_2}{1\text{ mol CaCl}_2}=191.6\text{ g CaCl}_2.

The answer would be that we're producing a mass of 191.6 g of CaCl₂.

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