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3.3 Yield and atom economy of chemical reactions

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Question 20

The balanced equation for a process at a chemical plant is shown.

2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}2NaOH+H2​SO4​→Na2​SO4​+2H2​O

For this process, sodium sulfate is the desired product.

Relative atomic masses: H=1\text{H} = 1H=1, O=16\text{O} = 16O=16, Na=23\text{Na} = 23Na=23, S=32\text{S} = 32S=32.

Which expression correctly calculates the percentage atom economy of the process?

[1]
A

2(2(1)+16)2(23+16+1)+(2(1)+32+4(16))×100\dfrac{2(2(1)+16)}{2(23+16+1)+(2(1)+32+4(16))}\times 1002(23+16+1)+(2(1)+32+4(16))2(2(1)+16)​×100

B

2(23)+32+4(16)(23+16+1)+(2(1)+32+4(16))×100\dfrac{2(23)+32+4(16)}{(23+16+1)+(2(1)+32+4(16))}\times 100(23+16+1)+(2(1)+32+4(16))2(23)+32+4(16)​×100

C

(2(23)+32+4(16))+2(2(1)+16)2(23+16+1)+(2(1)+32+4(16))×100\dfrac{(2(23)+32+4(16))+2(2(1)+16)}{2(23+16+1)+(2(1)+32+4(16))}\times 1002(23+16+1)+(2(1)+32+4(16))(2(23)+32+4(16))+2(2(1)+16)​×100

D

2(23)+32+4(16)2(23+16+1)+(2(1)+32+4(16))×100\dfrac{2(23)+32+4(16)}{2(23+16+1)+(2(1)+32+4(16))}\times 1002(23+16+1)+(2(1)+32+4(16))2(23)+32+4(16)​×100

Markscheme

3.3 Yield and atom economy of chemical reactions Questions

  1. GCSE
  2. /Chemistry
  3. /3.3 Yield and atom economy of chemical reactions

104 exam-style questions on AQA GCSE Chemistry 3.3 Yield and atom economy of chemical reactions, covering 3.3.1a Percentage yield, 3.3.1b Theoretical mass of a product, 3.3.2a Atom economy, and 3.3.2b Choosing a reaction pathway. Each one has a worked solution and a mark scheme showing where the marks go.

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