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SashulF [63]
3 years ago
14

A mixture contains 25 g of cyclohexane (C6H12) and 44 g of 2-methylpentane (C6H14). The mixture of liquids is at 35 oC . At this

temperature, the vapor pressure of pure cyclohexane is 150 torr, and that of pure 2-methylpentane is 313 torr. Assume this is an ideal solution. What is the mole fraction of cyclohexane in the liquid phase?
Chemistry
1 answer:
Greeley [361]3 years ago
5 0

Answer:

The mol fraction of cyclohexane in the liquid phase is 0.368

Explanation:

Step 1: Data given

Mass of cyclohexane = 25.0 grams

Mass of 2-methylpentane = 44.0 grams

Temperature = 35.0 °C

The pressure of cyclohexane = 150 torr

The pressure of 2-methylpentane = 313 torr

The pressure we only need for the mole fraction in gas phase.

Step 2: Calculate moles of cyclohexane

Moles cyclohexane = mass cyclohexane / molar mass

Moles cyclohexane = 25.0 g / 84 g/mol = 0.298 mol of cyclohexane

Step 3: Calculate moles of 2-methylpentane

Moles = 44.0 grams / 86 g/mol = 0.512 mol of 2-methylpentane

Step 4: Calculate mole fraction of cyclohexane in the liquid phase

Mole fraction of C6H12:

0.298 / (0.298 + 0.512) = 0.368

The mol fraction of cyclohexane in the liquid phase is 0.368

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N₂ : limiting reactant

H₂ : excess reactant

<h3>Further explanation</h3>

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

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Reaction

<em>N₂+3H₂⇒2NH₃</em>

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\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571

mol H₂(MW= 2 g/mol) :

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A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants

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