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luda_lava [24]
3 years ago
13

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by

12.0 torr at 35∘C. The vapor pressure of pure ethanol at 35∘C is 1.00×102 torr.
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

183.7 g of ethylene glycol

Explanation:

This a case of colligative property, about vapour pressure lowering.

ΔPv = P° . Xst

ΔPv = 12 Torr

P° = Vapour pressure of pure solvent → 100 Torr

12  Torr = 100 Torr . Xst

12 Torr / 100 Torr = 0.12 → Xst

Xst is mole fraction for solute which means moles of solute / Total moles and, the total moles are the sum of moles of solvent + moles of solute.

We can make this equation:

Moles of solute / (Moles of solute + Moles of solvent) = 0.12

We don't know the moles of solute, but we do know the moles of solvent by the mass.

Mass / Molar mass = Mol

1000 g / 46 g/m = 21.7 moles

Moles of solute / Moles of solute + 21.7 moles = 0.12

Moles of solute = 0.12 ( Moles of solute + 21.7 moles)

Moles of solute - 0.12 moles of solute = 2.60  moles

0.88 moles of solute = 2.60 moles

Moles of solute = 2.60 moles / 0.88 → 2.96 moles of solute

Mol . molar mass = Mass

2.96 m  .  62g/m = 183.7 g

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Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
3 years ago
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