Question 5.4: Additional Answer Options

2021 - May/June - Paper 2 - Question 5

Or

Option 2

\text{ ave rate }=\frac{\Delta\mathrm{n}}{\Delta\mathrm{t}}

4,4\times10^{-3}=\frac{\Delta n}{2,5}

\Delta\mathrm{n}\left[\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3\right]=0,016-0,011

=0.005\mathrm{~mol}

Or

Option 3

\text{ ave. rate }=\frac{\Delta\mathrm{n}}{\Delta\mathrm{t}}

4,4\times10^{-3}=\frac{\Delta n}{2,5}

\Delta\mathrm{n}\left(\mathrm{CO}_2\right)=0,011\mathrm{~mol}

\mathrm{n}\left(\mathrm{CO}_2\right):\mathrm{n}\left(\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3\right.

3:1

0,011:3,67\times10^{-3}\mathrm{~mol}

\mathrm{n}\left(\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3\text{ left }=0,016-3,67\times10^{-3}=1,23\times10^{-2}\mathrm{mol}\right.

Or

Option 4

\text{ ave. rate }=\frac{\Delta\mathrm{n}}{\Delta\mathrm{t}}

4,4\times10^{-3}=\frac{\Delta n}{2,5}

\Delta\mathrm{n}(\mathrm{HCl})=0,011\mathrm{~mol}

\mathrm{n}\left[\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3\right]=\frac{0,011}{6}=0,0018\mathrm{~mol}

\mathrm{n}\left[\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3\right]\text{ left }=0,016-0,0018=0,0142\mathrm{mol}

Or

Option 5

\text{ ave. rate}=\frac{\Delta\mathrm{n}}{\Delta\mathrm{t}}

4,4\times10^{-3}=\frac{\Delta n}{2,5}

\Delta\mathrm{n}\left(\mathrm{AlCl}_3\right)=0,011\mathrm{~mol}

\mathrm{n}\left[\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3\right]=0,0055\mathrm{~mol}

\mathrm{n}\left[\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3\right]\text{ left}=0,016-0,0055=0,0105\mathrm{mol}

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Physical Sciences
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