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The complete combustion of 1.5 x 10^3 moles of a gaseous hydrocarbon required 84 cm^3 of oxygen (measured as 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 cm^3 of oxygen (measured as 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 cm^3 of oxygen (measured as 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 calculate the number of moles of oxygen and water produced, we start by using the ideal gas law and the conditions at standard temperature and pressure (S.T.P.).

Moles of Oxygen:

Given the volume of oxygen needed for combustion: 84 cm³. The molar volume of a gas at S.T.P. is 22,400 cm³/mol.

$ext{Moles of } O_2 = \frac{84 \text{ cm}^3}{22400 \text{ cm}^3/mol} = 0.00375 \text{ mol} \approx 2.5 \text{ mol} \text{ for } \text{1.5 x } 10^3 moles$
Moles of Water:

Given that 27 mg of water is produced:
- Convert mg to g:
  
  $27 ext{ mg} = 0.027 ext{ g}$
- We know the molar mass of water (H2O) is approximately 18 g/mol:
  
  $ext{Moles of } H_2O = \frac{0.027 ext{ g}}{18 ext{ g/mol}} \approx 0.0015 ext{ mol} \approx 1.5 \text{ for 1 mole of hydrocarbon}$

Step 2

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

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Answer

To verify that the gaseous hydrocarbon is ethyne, we can use the stoichiometric relationships from the combustion reaction:

Combustion of ethyne:

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

From earlier calculations:

Moles of $O_2$ used = $2.5$ moles
Moles of $CO_2$ produced = $2$ moles (from combustion)

Using the empirical formula approach:

The ratio of moles of carbon to moles of hydrogen in ethyne is:

$C_2H_y \Rightarrow (2 : 2)$

From the empirical formula, we confirm that the gaseous hydrocarbon is indeed C2H2, which is 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 (also known as ethyl alcohol), which has the chemical formula C2H5OH.

Reagents Required:

Dilute sulphuric acid (H2SO4)
Water (H2O)

Conditions Required:

The reaction typically occurs at room temperature or can be warmed gently to facilitate the process.
Alternatively, the reaction can involve a stronger acid catalyst, but dilute sulphuric acid is commonly sufficient for hydration.

The complete combustion of 1.5 x 10^3 moles of a gaseous hydrocarbon required 84 cm^3 of oxygen (measured as 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 cm^3 of oxygen (measured as 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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