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The complete combustion of 1.5 x 10^-3 moles of a gaseous hydrocarbon required 84 cm³ of oxygen (measured at S.T.P) and produced 27 mg of water - Leaving Cert Chemistry - Question b - 2001

Question b

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The complete combustion of 1.5 x 10^-3 moles of a gaseous hydrocarbon required 84 cm³ of oxygen (measured at S.T.P) and produced 27 mg of water. (i) How many moles ... show full transcript

Worked Solution & Example Answer:The complete combustion of 1.5 x 10^-3 moles of a gaseous hydrocarbon required 84 cm³ of oxygen (measured at S.T.P) and produced 27 mg of water - Leaving Cert Chemistry - Question b - 2001

Step 1

How many moles of oxygen would be used up and how many moles of water would be produced if one mole of the hydrocarbon were burned in oxygen.

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Answer

To find the moles of oxygen used, we'll use the ideal gas law. From the problem, the volume of oxygen is 84 cm³, which we convert to liters:

$84 ext{ cm}^3 = 0.084 ext{ L}$

Using the molar volume of a gas at STP, which is 22.4 L/mol, we calculate:

$ext{Moles of } O_2 = \frac{0.084 ext{ L}}{22.4 ext{ L/mol}} \approx 0.00375 ext{ moles}$

For the moles of water produced, we convert the mass of water (27 mg) to grams:

$27 ext{ mg} = 0.027 ext{ g}$

Now, using the molar mass of water (18 g/mol):

$ext{Moles of } H_2O = \frac{0.027}{18} \approx 0.0015 ext{ moles}$

Therefore, for every mole of hydrocarbon burned, approximately 0.00375 moles of oxygen are used, producing 0.0015 moles of water.

Step 2

Show clearly that the gaseous hydrocarbon is ethyne (C₂H₂).

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Answer

To establish that the hydrocarbon is ethyne (C₂H₂), we can analyze the combustion reaction:

For the combustion of C₂H₂:

$C_2H_2 + 2 O_2 \rightarrow 2 CO_2 + H_2O$

Using the stoichiometry:

2 moles of oxygen are needed to combust 1 mole of C₂H₂. Given: 0.00375 moles of O₂ → Using the ratio: $\frac{1 ext{ mol } C_2H_2}{2 ext{ mol } O_2}$ We calculate moles of C₂H₂: $\text{Moles of } C_2H_2 = \frac{0.00375}{2} = 0.001875 ext{ moles}$

Next, calculating moles of produced CO₂ and H₂O: Given total moles of products:

Moles of CO₂ produced = 0.001875 imes 2 = 0.00375 moles.
Moles of water produced = 0.001875 moles.

Carbon to hydrogen ratio confirms the empirical formula is C₂H₂, demonstrating the hydrocarbon is indeed ethyne.

Step 3

What is the product of the hydration of ethyne? What reagents and conditions are required for this hydration?

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Answer

The product of the hydration of ethyne (C₂H₂) is ethanol (C₂H₅OH) or acetaldehyde.

Reagents and Conditions:

Reagents: The hydration reaction typically requires water (H₂O) and a catalyst such as dilute sulfuric acid (H₂SO₄) or mercury(II) sulfate (HgSO₄).
Conditions: The reaction conditions involve heating at temperatures generally between 40°C and 100°C, with sufficient pressure to favor the addition reaction.

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The complete combustion of 1.5 x 10^-3 moles of a gaseous hydrocarbon required 84 cm³ of oxygen (measured at S.T.P) and produced 27 mg of water - Leaving Cert Chemistry - Question b - 2001

Worked Solution & Example Answer:The complete combustion of 1.5 x 10^-3 moles of a gaseous hydrocarbon required 84 cm³ of oxygen (measured at S.T.P) and produced 27 mg of water - Leaving Cert Chemistry - Question b - 2001

How many moles of oxygen would be used up and how many moles of water would be produced if one mole of the hydrocarbon were burned in oxygen.

Show clearly that the gaseous hydrocarbon is ethyne (C₂H₂).

What is the product of the hydration of ethyne? What reagents and conditions are required for this hydration?

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