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A student carried out an investigation to determine the order of reactivity of four metals, W, X, Y and Z - Edexcel - GCSE Chemistry - Question 9 - 2022 - Paper 1

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A student carried out an investigation to determine the order of reactivity of four metals, W, X, Y and Z. A piece of metal W was added to a test tube containing ex... show full transcript

Worked Solution & Example Answer:A student carried out an investigation to determine the order of reactivity of four metals, W, X, Y and Z - Edexcel - GCSE Chemistry - Question 9 - 2022 - Paper 1

Step 1

Use the information in Figure 8 to place the metals in order of reactivity from the least reactive to the most reactive.

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Answer

Based on the observations:

  • W shows some reaction but leaves metal remaining, indicating it is less reactive than others.
  • X demonstrates minimal reaction with no change, so it is more reactive than W but less than Y and Z.
  • Y reacts completely, making it more reactive than W and X.
  • Z reacts quickly with no metal remaining, placing it as the most reactive.

Thus, the order is: W < X < Y < Z.

Step 2

Explain why the reaction between metal Y and excess dilute sulfuric acid stopped even though there was solid metal Y left.

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Answer

The reaction between metal Y and dilute sulfuric acid would stop due to the formation of a protective layer of sulfate on the surface of metal Y. This layer inhibits further reaction because it prevents the acid from coming into contact with the unreacted metal, halting the reaction despite the presence of excess acid.

Step 3

Explain the meaning of the term weak acid.

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Answer

A weak acid is one that does not completely dissociate in solution. This means that only a fraction of the acid molecules release hydrogen ions into the solution, resulting in a lower concentration of hydrogen ions compared to strong acids. Consequently, weak acids exhibit weaker acidic properties and slower reaction rates with metals compared to stronger acids.

Step 4

Calculate the total number of atoms that combine to form 5.13 g of aluminium sulfate.

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Answer

First, calculate the molar mass of Al₂(SO₄)₃:

extMolarmass=2imes27.0+3imes(32.0+4imes16.0)=342.0extg/mol ext{Molar mass} = 2 imes 27.0 + 3 imes (32.0 + 4 imes 16.0) = 342.0 ext{ g/mol}

Next, find the number of moles in 5.13 g:

extMolesofAl2(SO4)3=5.13extg342.0extg/mol=0.015extmoles ext{Moles of Al₂(SO₄)₃} = \frac{5.13 ext{ g}}{342.0 ext{ g/mol}} = 0.015 ext{ moles}

The formula of aluminium sulfate contains:

  • 2 Aluminium (Al)
  • 3 Sulfur (S)
  • 12 Oxygen (O)

Total atoms per formula unit = 2 + 3 + 12 = 17.

Thus, total number of atoms in 5.13 g:

ightarrow 1.53 imes 10^{19} ext{ atoms}$$

Step 5

Carry out a calculation to show which equation represents the reaction taking place.

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Answer

First, determine the moles of iron (Fe) used:

extMolesofFe=4.48extg56.0extg/mol=0.08extmoles ext{Moles of Fe} = \frac{4.48 ext{ g}}{56.0 ext{ g/mol}} = 0.08 ext{ moles}

In Equation 1, the molar ratio is 1:1, so we need 0.08 moles of Pb, which would give:

0.08extmolesofPbimes207extg/mol=16.56extgofPb0.08 ext{ moles of Pb} imes 207 ext{ g/mol} = 16.56 ext{ g of Pb}

In Equation 2, the ratio is 2:3. Therefore, for 0.08 moles of Fe:

32imes0.08=0.12extmolesofPbrequired \frac{3}{2} imes 0.08 = 0.12 ext{ moles of Pb required}

This translates to:

0.12imes207=14.52extgofPb0.12 imes 207 = 14.52 ext{ g of Pb}

Since 24.84 g of Pb is produced, it suggests that Equation 2 with a larger yield aligns with the experimental observation, hence it is the likely equation representing the reaction.

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