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The diagram shows a piece of magnesium ribbon being heated - Edexcel - GCSE Chemistry - Question 4 - 2012 - Paper 1

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The diagram shows a piece of magnesium ribbon being heated. During the heating, the magnesium reacts with oxygen from the air. The lid of the crucible was raised sl... show full transcript

Worked Solution & Example Answer:The diagram shows a piece of magnesium ribbon being heated - Edexcel - GCSE Chemistry - Question 4 - 2012 - Paper 1

Step 1

4(b)(i) Plot the data correctly on a graph.

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Answer

To plot the data correctly, take the mass of magnesium used as the x-axis and the mass of magnesium oxide produced as the y-axis. Each point represents the corresponding mass of magnesium and the respective mass of magnesium oxide formed from the table. Ensure to label the axes accurately and include a suitable scale. Draw a line of best fit through the points to indicate the relationship.

Step 2

4(b)(ii) Explain why not all magnesium was reacted.

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Answer

One possible reason that not all of the magnesium was reacted is that the lid of the crucible was not lifted for a sufficient amount of time, preventing enough oxygen from entering to fully react with the magnesium. Additionally, some magnesium might have been left unburned or there could have been smoke loss, which means that some magnesium oxide may have escaped during the experiment.

Step 3

4(c) Write a balanced equation for the reaction.

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The balanced chemical equation for the reaction between magnesium and oxygen is:

ightarrow 2MgO$$ This equation shows that two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide.

Step 4

4(d) Determine the empirical formula of magnesium oxide.

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To find the empirical formula, we analyze the ratio of magnesium to oxygen in the reactants. The molar mass of magnesium (Mg) is approximately 24.31 g/mol and for oxygen (O) is approximately 16.00 g/mol. From the data:

  1. For experiment 1:
    • Mass of Mg = 0.10 g
    • Mass of O in MgO = 0.06 g
    • Moles of Mg = 0.10 g / 24.31 g/mol = 0.00412 moles
    • Moles of O = 0.06 g / 16.00 g/mol = 0.00375 moles
    • Ratio = 0.00412 : 0.00375 = 1.1 : 1
    • Empirical formula: approximately MgOMgO.

Repeat for other experiments to confirm the ratio is consistent.

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