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ivolga24 [154]
3 years ago
12

1‑propanol ( n ‑propanol) and 2‑propanol (isopropanol) form ideal solutions in all proportions. Calculate the partial pressure a

nd the mole fraction ( y ) of the vapor phase of each component in equilibrium with each of the given solutions at 25 °C. P ∘ prop = 20.9 Torr and P ∘ iso = 45.2 Torr at 25 °C. A solution with a mole fraction of x prop = 0.247 .
Chemistry
1 answer:
Alex777 [14]3 years ago
5 0

Answer:

y_{prop} = 0.134; y_{iso} = 0.866

The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr

Explanation:

For each of the solutions:

mole fraction of isopropanol  (x_{iso}) = 1 - mole fraction of propanol (x_{prop}).

Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.

Furthermole, the partial pressure of isopropanol = x_{iso}*vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr

The partial pressure of propanol = x_{prop}*vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr

Similarly,

In the vapor phase,

The mole fraction of propanol (y_{prop}) = \frac{P_{prop} }{P_{prop}+P_{iso}}

Where, P_{prop} is the partial pressure of propanol and P_{iso} is the partial pressure of isopropanol.

Therefore,

y_{prop} = 5.26/(34.04+5.16) = 0.134

y_{iso} = 1 - 0.134 = 0.866

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Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of c
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The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

Answer:

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Explanation:

The equation of the reaction:

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Amount of potassium phosphate= 500/212.27= 2.4 moles

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Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

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b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

d)

3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

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