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a) Methane is a hydrocarbon fuel - Edexcel - GCSE Chemistry - Question 6 - 2020 - Paper 1

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a) Methane is a hydrocarbon fuel. (i) Complete the word equation for the complete combustion of methane in oxygen: methane + oxygen → water (ii) The incomplete co... show full transcript

Worked Solution & Example Answer:a) Methane is a hydrocarbon fuel - Edexcel - GCSE Chemistry - Question 6 - 2020 - Paper 1

Step 1

Complete the word equation for the complete combustion of methane in oxygen:

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Answer

The complete word equation for the complete combustion of methane in oxygen is:

 methane + oxygen → water + carbon dioxide

Step 2

Give the reason why carbon and carbon monoxide are produced in the incomplete combustion of methane:

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Answer

Carbon and carbon monoxide are produced in the incomplete combustion of methane due to a limited supply of oxygen. When oxygen is insufficient, not all carbon atoms are converted to carbon dioxide, leading to the formation of solid carbon (soot) and carbon monoxide (a toxic gas).

Step 3

Name the fraction in Figure 10 that is used to surface roads:

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Answer

The fraction in Figure 10 that is used to surface roads is "bitumen."

Step 4

Name the fraction in Figure 10 that contains hydrocarbons with the lowest boiling point:

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Answer

The fraction in Figure 10 that contains hydrocarbons with the lowest boiling point is "gases."

Step 5

Which of the following shows the fractions where the relative demand is greater than the relative amount in the crude oil?

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Answer

The correct answer is B) gases, petrol, diesel oil.

Step 6

Determine the value of x in the molecule of C₆H₋ₓ:

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Answer

In the cracking reaction, octane (C₈H₁₈) produces one molecule of ethane (C₂H₆) and one molecule of C₆H₋ₓ. The total number of carbon atoms must remain the same:

eight (from octane) = two (from ethane) + six (from C₆H₋ₓ).

Thus, x = 6.

Step 7

Calculate the maximum mass of octane that could be produced when 340g of dodecane is cracked in this reaction:

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Answer

To find the maximum mass of octane produced, first establish the number of moles of dodecane:

extNumberofmolesofdodecane=340g170g/mol=2.0extmoles ext{Number of moles of dodecane} = \frac{340g}{170g/mol} = 2.0 ext{ moles}

Since one mole of dodecane yields one mole of octane, the number of moles of octane produced will equal the number of moles of dodecane:

extMolesofoctane=2.0extmoles ext{Moles of octane} = 2.0 ext{ moles}

The mass of octane can be calculated as follows:

extMassofoctane=Moles of octane×Molar mass of octane=2.0extmoles×114g/mol=228g ext{Mass of octane} = \text{Moles of octane} \times \text{Molar mass of octane} = 2.0 ext{ moles} \times 114g/mol = 228g

Therefore, the maximum mass of octane that could be produced is 228g.

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