Question 5.4: Additional Answer Options

2021 - November - Paper 2 - Question 5

Or

Option 2

\mathrm{n}\left(\mathrm{CaCO}_3\right)=\frac{\mathrm{m}}{\mathrm{M}}

=\frac{15}{100}

=0,15\mathrm{~mol}

\mathrm{n}\left(\mathrm{CO}_2\right)=\mathrm{n}\left(\mathrm{CaCO}_3\right)

=0,15\mathrm{~mol}

\mathrm{n}\left(\mathrm{CO}_2\right)=\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{M}}}

0,15=\frac{V}{24000}\quad/0,15=\frac{V}{24}

V=3600\mathrm{~cm}^3,\quad V=3,6\mathrm{dm}^3

\mathrm{V}\left(\mathrm{CO}_2\right)=\frac{82,5}{100}\times3600/3,6

=2976\mathrm{~cm}^3/2,976\mathrm{dm}^3

Or

Option 3

=\frac{15}{100}

=0,15\mathrm{~mol}

\mathrm{n}\left(\mathrm{CO}_2\right)=\mathrm{n}\left(\mathrm{CaCO}_3\right)

=0,15\mathrm{~mol}

\mathrm{n}\left(\mathrm{CO}_2\right)=\frac{\mathrm{m}}{\mathrm{M}}

\mathrm{m}\left(\mathrm{CO}_2\right)=0,15\times44

=6,6\mathrm{~g}

82,5=\frac{m_{\text{actual}}}{6,6}\times100

m_{(\text{actual \rparen}}=5,445\mathrm{g}

n\left(C_2\right)=\frac{m}{M}

=\frac{5,445}{44}

=0,12375\mathrm{~mol}

\mathrm{n}\left(\mathrm{CO}_2\right)=\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{M}}}

0,12375=\frac{V}{24000}OR\quad0,12375=\frac{V}{24}

V=2976\mathrm{~cm}^3/2,976\mathrm{dm}^3

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Physical Sciences
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