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This question is about metals and metal compounds - AQA - GCSE Chemistry - Question 2 - 2018 - Paper 1

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This question is about metals and metal compounds. 02.1 Iron pyrites is an ionic compound. Figure 1 shows a structure for iron pyrites. Determine the formula of ... show full transcript

Worked Solution & Example Answer:This question is about metals and metal compounds - AQA - GCSE Chemistry - Question 2 - 2018 - Paper 1

Step 1

Determine the formula of iron pyrites.

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Answer

The formula for iron pyrites is FeS2FeS_2. This compound consists of one iron (Fe) atom and two sulfur (S) atoms.

Step 2

Number of protons

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Answer

The number of protons in an atom of iron is 26.

Step 3

Number of neutrons

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Answer

The number of neutrons in an atom of iron is 30.

Step 4

Number of electrons

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Answer

The number of electrons in an atom of iron is 26.

Step 5

Give two differences between the properties of iron and sodium.

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Answer

  1. Iron has a higher melting and boiling point than sodium, making it more suitable for structural applications.
  2. Iron is stronger and denser compared to sodium, which is softer and lighter.

Step 6

Explain why carbon can be used to extract nickel from nickel oxide.

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Answer

Carbon can be used to extract nickel from nickel oxide because it is more reactive than nickel. During the reduction process, carbon displaces nickel from nickel oxide by removing oxygen, thereby allowing the nickel to be recovered.

Step 7

Calculate the percentage atom economy for the reaction to produce nickel.

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Answer

To calculate the percentage atom economy:

  1. Determine the total molar mass of the reactants:
    Molar mass of NiO = 75 g/mol + Molar mass of C = 12 g/mol = 87 g/mol.

  2. The molar mass of the desired product (nickel) is 59 g/mol.

  3. Apply the atom economy formula:

    extPercentageAtomEconomy=(Molar mass of desired productTotal molar mass of reactants)×100 ext{Percentage Atom Economy} = \left( \frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \right) \times 100

    =(5987)×10067.68%= \left( \frac{59}{87} \right) \times 100 \approx 67.68\%

Thus, the percentage atom economy is approximately 67.7% when rounded to three significant figures.

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