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Citrus2011 [14]
3 years ago
8

Separation units often use membranes, absorbers, and other devices to reduce the mole fraction of selected con- stituents in gas

eous mixtures. Consider a mixture of hydro- carbons that consists of 60 percent (by volume) methane, 30 percent ethane, and 10 percent propane. After passing through a separator, the mole fraction of the propane is reduced to 1 percent. The mixture pressure before and after the separa- tion is 100 kPa. Determine the change in the partial pressures of all the constituents in the mixture
Chemistry
1 answer:
Viktor [21]3 years ago
8 0

Answer:

ΔP(ch4) = 6kPa

ΔP(c2h6) = 3kPa

ΔP(c3h8) = -9kPa

Explanation:

first, we find the factor that is used in the determination of mole fraction of the other components using the mole fraction of propane, both at the beginning and at the end. This means

β/(0.6+0.3+β) = 0.01

β = 0.009 + 0.01β

β - 0.01β = 0.009

0.99β = 0.009

β = 0.00909

the mole fractions of the methane and ethane at the end are

yCH4 = 0.6/(0.6 + 0.3 +0.00909)

yCH4 = 0.6/0.90909

yCH4 = 0.66

yC2H6 = 0.3/(0.6 + 0.3 + 0.00909

yC2H6 = 0.3/0.90909

yC2H6 = 0.33

Now, the change in partial pressure will be determined from the mole fraction at the beginning and at the end

ΔP(ch4) = (y2 - y1)ch4 * P

ΔP(ch4) = (0.66 - 0.6) * 100

ΔP(ch4) = 6kPa

ΔP(c2h6) = (y2 - y1)c2h6 * P

ΔP(c2h6) = (0.33 - 0.3) * 100

ΔP(c2h6) = 3kPa

ΔP(c3h8) = (y2 - y1)c3h8 * P

ΔP(c3h8) = (0.01 - 0.1) * 100

ΔP(c3h8) = -9kPa

Therefore

ΔP(ch4) = 6kPa

ΔP(c2h6) = 3kPa

ΔP(c3h8) = -9kPa

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Answer:

0.49 mol

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCI ⇒ MgCl₂ + H₂

Step 2: Calculate the moles corresponding to 12 g of Mg

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NH_4I(s)\rightarrow NH_3(g)+HI(g)

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Explanation:

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A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.

1. N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products,  the randomness is decreasing and the entropy is negative

2. 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

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3. NH_4I(s)\rightarrow NH_3(g)+HI(g)

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5. 2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)

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Answer:

a)  No molecules of hydrogen

b) four molecules of ammonia

c) four left molecules of nitrogen.

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The balanced reaction between nitrogen and hydrogen molecules to give ammonia molecules is:

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Thus one molecule of nitrogen will react with three molecules of hydrogen to give two molecules of ammonia.

We have six molecules of each nitrogen and hydrogen in the closed container and they undergo complete reaction it means the limiting reagent is hydrogen. For six molecules of nitrogen, eighteen molecules of hydrogen will be required.

So six molecules of hydrogen will react with two molecules of nitrogen to give four molecules of ammonia.

The product mixture will have

a) No molecules of hydrogen

b) four molecules of ammonia

c) four left molecules of nitrogen.

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