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nirvana33 [79]
3 years ago
11

Problem 12.002 the molar analysis of a gas mixture at 30°c, 2 bar is 40% n2, 50% co2, 10% ch4. determine

Chemistry
1 answer:
andreev551 [17]3 years ago
3 0
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.

1.

Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar

2. For the volume, let's find the total volume first.

V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
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Answer : The mass of potassium hypochlorite is, 4.1 grams.

Explanation : Given,

pH = 10.20

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KClO\rightarrow K^++ClO^-

Now the further reaction with water (H_2O) to give,

ClO^-+H_2O\rightarrow HClO+OH^-

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Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

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[OH^-]=1.58\times 10^{-4}M

Now we have to calculate the base dissociation constant.

Formula used : K_b=\frac{K_w}{K_a}

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K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}

Now we have to calculate the concentration of ClO^-.

The equilibrium constant expression of the reaction  is:

K_b=\frac{[OH^-][HClO]}{[ClO^-]}

As we know that, [OH^-]=[HClO]=1.58\times 10^{-4}M

2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}

[ClO^-]=0.0999M

Now we have to calculate the moles of ClO^-.

\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}

\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole

As we know that, the number of moles of ClO^- are equal to the number of moles of KClO.

So, the number of moles of KClO = 0.0449 mole

Now we have to calculate the mass of KClO.

\text{Mass of }KClO=\text{Moles of }KClO\times \text{Molar mass of }KClO

\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g

Therefore, the mass of potassium hypochlorite is, 4.1 grams.

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