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This question is about silver iodide - AQA - GCSE Chemistry - Question 3 - 2020 - Paper 1

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This question is about silver iodide. Silver iodide is produced in the reaction between silver nitrate solution and sodium iodide solution. The equation for the re... show full transcript

Worked Solution & Example Answer:This question is about silver iodide - AQA - GCSE Chemistry - Question 3 - 2020 - Paper 1

Step 1

Explain how the results demonstrate the law of conservation of mass.

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Answer

To demonstrate the law of conservation of mass, we need to recognize that the total mass before and after the reaction should be equivalent. The total mass before the reaction is calculated as:

78.26+78.50=156.76extg78.26 + 78.50 = 156.76 ext{ g}

After mixing, the total mass becomes:

108.22+48.54=156.76extg108.22 + 48.54 = 156.76 ext{ g}

Since both totals are the same, it shows that no mass was lost or gained during the chemical reaction.

Step 2

Suggest how the student could separate the insoluble silver iodide from the mixture at the end of the reaction.

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Answer

The student could use filtration to separate the insoluble silver iodide from the mixture. This would involve pouring the mixture through filter paper in a funnel, where the solid silver iodide would remain on the paper while the liquid passes through.

Step 3

Suggest one impurity that was removed by rinsing with water.

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Answer

One impurity that was likely removed by rinsing with water could be any soluble salts, such as sodium nitrate, which are soluble in water.

Step 4

Suggest why the student warmed the silver iodide.

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Answer

The student warmed the silver iodide to help remove any remaining water and to potentially make it easier to handle or to ensure it is dry for further reactions or analysis.

Step 5

Calculate the percentage atom economy for the production of silver iodide in this reaction.

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Answer

The total molar mass of reactants is:

Mr=170+150=320M_r = 170 + 150 = 320

The molar mass of the desired product is:

Mr(extAgI)=235M_r ( ext{AgI}) = 235

The percentage atom economy is calculated as:

extPercentageAtomEconomy=(235320)×100=73.4375% ext{Percentage Atom Economy} = \left(\frac{235}{320}\right) \times 100 = 73.4375\%

Rounding to three significant figures gives:

73.4%73.4\%

Step 6

Give one reason why reactions with a high atom economy are used in industry.

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Answer

Reactions with a high atom economy are preferred in industry because they generate less waste, leading to more sustainable processes and cost savings in waste disposal.

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