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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 equ... 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 is potassium chloride, commonly abbreviated as KCl.

Step 2

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

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Answer

The ionic equation for the neutralisation is: H++OHH2OH^+ + OH^- → 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

The correct choice is 'Copper carbonate and copper oxide only'.

Step 4

Give one reason for: step 2

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Answer

In step 2, warming the sulfuric acid is done to speed up the reaction rate between sulfuric acid and magnesium oxide.

Step 5

Give one reason for: step 5

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Answer

In step 5, this step ensures that all the hydrochloric acid reacts with the magnesium oxide to form magnesium sulfate.

Step 6

Give one reason for: step 6

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Answer

In step 6, filtering the mixture is necessary to remove any excess magnesium oxide from the solution.

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 avoid overheating and breaking the crystals.

Step 8

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

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Answer

To calculate the volume of chlorine needed, first determine the moles of iron: extmolesFe=14extg56extg/mol=0.25extmol ext{moles Fe} = \frac{14 ext{ g}}{56 ext{ g/mol}} = 0.25 ext{ mol}

From the balanced equation: 2extFe+3extCl22extFeCl32 ext{ Fe} + 3 ext{ Cl}_2 \rightarrow 2 ext{ FeCl}_3

Thus, for 0.25 moles of Fe: extmolesCl2=32×0.25=0.375extmol ext{moles Cl}_2 = \frac{3}{2} \times 0.25 = 0.375 ext{ mol}

Finally, calculate the volume of chlorine gas needed: extvolumeCl2=24extdm3imes0.375=9.0extdm3 ext{volume Cl}_2 = 24 ext{ dm}^3 imes 0.375 = 9.0 ext{ dm}^3

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