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This question is about salts - AQA - GCSE Chemistry - Question 5 - 2021 - Paper 1

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This question is about salts. Name the salt produced by the neutralisation of hydrochloric acid with potassium hydroxide. Write an ionic equation for the neutralis... show full transcript

Worked Solution & Example Answer:This question is about salts - AQA - GCSE Chemistry - Question 5 - 2021 - Paper 1

Step 1

Name the salt produced by the neutralisation of hydrochloric acid with potassium hydroxide.

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Answer

The salt produced by the neutralization is potassium chloride (KCl).

Step 2

Write an ionic equation for the neutralisation of hydrochloric acid with potassium hydroxide.

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Answer

H++OHH2OH^+ + OH^- \rightarrow H_2O

Step 3

Which of these insoluble solids can be used to make a copper salt by reacting the solid with dilute hydrochloric acid?

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Answer

Copper carbonate and copper oxide only.

Step 4

Give one reason for: step 2

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Answer

Step 2 (to speed up the reaction) involves warming the sulfuric acid, which can increase the reaction rate by providing more energy to the particles involved.

Step 5

Give one reason for: step 5

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Answer

Step 5 is to ensure that all the hydrochloric acid reacts with magnesium oxide, preventing any unreacted acid from remaining in the solution.

Step 6

Give one reason for: step 6

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Answer

Step 6 is to remove the excess magnesium oxide, which is accomplished by filtering the mixture.

Step 7

How should the filtrate be evaporated gently in step 7?

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Answer

The filtrate should be evaporated gently using a boiling water bath or an electric heater to prevent the crystals from forming too quickly.

Step 8

Calculate the volume of chlorine needed to react with 14 g of iron.

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Answer

First, calculate the moles of iron: moles Fe=14g56g/mol=0.25 mol\text{moles Fe} = \frac{14 g}{56 g/mol} = 0.25 \text{ mol}. Next, from the reaction equation, 2 moles of Fe react with 3 moles of Cl₂, meaning 0.25 moles of iron would require 0.375 moles of chlorine: moles Cl₂=0.25 mol×32=0.375 mol\text{moles Cl₂} = 0.25 \text{ mol} \times \frac{3}{2} = 0.375 \text{ mol}. Finally, the volume of chlorine needed is: Volume Cl₂=24dm3/mol×0.375mol=9.0dm3\text{Volume Cl₂} = 24 dm³/mol \times 0.375 mol = 9.0 dm³.

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