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Magnesium carbonate has the formula MgCO₃ - Edexcel - GCSE Chemistry - Question 3 - 2022 - Paper 1

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Magnesium carbonate has the formula MgCO₃. (a) Magnesium carbonate contains Mg²⁺ and CO₃²⁻ ions. (i) The atomic number of magnesium is 12. What is the electronic ... show full transcript

Worked Solution & Example Answer:Magnesium carbonate has the formula MgCO₃ - Edexcel - GCSE Chemistry - Question 3 - 2022 - Paper 1

Step 1

What is the electronic configuration of the Mg²⁺ ion?

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Answer

The electronic configuration of magnesium (atomic number 12) is 2.8. When magnesium loses two electrons to form the Mg²⁺ ion, its electron configuration becomes 2, as the outermost shell (2.8) loses the two electrons.

Step 2

Explain why solid magnesium carbonate cannot conduct electricity but solid magnesium can.

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Solid magnesium can conduct electricity because it consists of metallic bonding with free-moving electrons. In contrast, solid magnesium carbonate is an ionic compound, and while the ions are present, they are held in fixed positions within the crystal lattice, preventing movement. Hence, solid magnesium carbonate cannot conduct electricity.

Step 3

Calculate the percentage by mass of magnesium in magnesium carbonate, MgCO₃.

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To find the percentage by mass of magnesium in MgCO₃, we first calculate the molar mass of the compound:

  • Molar mass of MgCO₃ = Mg (24.0) + C (12.0) + 3×O (3×16.0) = 24.0 + 12.0 + 48.0 = 84.0 g/mol.

Next, the percentage by mass of magnesium is given by the formula: Percentage by mass of Mg=(Mass of MgMolar Mass of MgCO₃)×100\text{Percentage by mass of Mg} = \left( \frac{\text{Mass of Mg}}{\text{Molar Mass of MgCO₃}} \right) \times 100

Substituting the values: (24.084.0)×100=28.57%\left( \frac{24.0}{84.0} \right) \times 100 = 28.57\%

Step 4

Complete the balanced equation for this reaction.

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The balanced equation for the reaction of magnesium carbonate with dilute hydrochloric acid is:

MgCO3+2HClMgCl2+H2O+CO2\text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O} + \text{CO}_2

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