Couple prelim questions (1 Viewer)

piethepker · May 19, 2015

Collectively as a class we couldn't figure these two out, could anyone help out?

40. If the theoretical yield of aluminium from a particular bauxite ore is 14% and if the extraction process is only 87% efficient, what mass of bauxite needs to be processed to produce 1.00 tonne of aluminium

41. What mass of sodium would be produced from the electrolysis of 1.00 tonne of molten sodium chloride if the electrolysis process was 92% efficient?

Answers

40 8.3 t
41 0.26 t

Cheers

InteGrand · May 19, 2015

piethepker said:
Collectively as a class we couldn't figure these two out, could anyone help out?

40. If the theoretical yield of aluminium from a particular bauxite ore is 14% and if the extraction process is only 87% efficient, what mass of bauxite needs to be processed to produce 1.00 tonne of aluminium

41. What mass of sodium would be produced from the electrolysis of 1.00 tonne of molten sodium chloride if the electrolysis process was 92% efficient?

Answers

40 8.3 t
41 0.26 t

Cheers

$40. Let the required mass of bauxite be $m_\text{bauxite}$, then the obtained yield of aluminium is given by: $\text{Obtained yield}=\text{Efficiency}\times \text{Theoretical percentage yield}\times m_\text{bauxite}\Rightarrow m_\text{bauxite} = \frac{\text{Obtained yield}}{\text{Efficiency}\times \text{Theoretical percentage yield}}$. Since we want $\text{Obtained yield} = 1.00 \text{ t}$ and have $\text{Efficiency} = 87\%, \text{Theoretical percentage yield} = 14\%$, we have $m_\text{bauxite} = \frac{1.00 \text{ t}}{0.87\times 0.14}=8\tfrac{128}{609}\text{ t}\approx8.2 \text{ t}$.$

$41. The chemical equation for the electrolysis is $2\text{NaCl}_{(\ell)}\to 2\text{Na}_{(s)}+\text{Cl}_2 _{(g)}$. So we can see that for each mole of NaCl used, we would theoretically obtain one mole of Na. But since the process is 92\% efficient, we actually obtain 0.92 mol of Na per mole of NaCl used.$

$Now, the molar mass of NaCl is 58.44 g/mol (this can be calculated from the Periodic Table). So in 1.00 tonne (i.e. $10^6$ g) of NaCl, the quantity of NaCl is $n=\frac{m}{MM}=\frac{10^6 \text{ g}}{58.44 \text{ g/mol} } = A\text{ mol}$, say ($A\approx 17111.57$, but we'll use exact values until the very end, when we'll round). So we obtain $0.92A$ mol of Na. Since the molar mass of Na is 22.9898 g/mol (I'm getting these values from Google, the values in the HSC Chemistry Data Page are slightly different), this means the mass of Na obtained is $m_{\text{Na}}=n_{\text{Na}}\times MM_{\text{Na}} = 0.92A \text{ mol}\times 22.9898 \text{ g/mol}\approx 361920.1916 \text{ g} =\frac{361920.1916}{10^6}\text{ t} \approx 0.36 \text{ t}.$$

piethepker · May 20, 2015

Ah so the answers were wrong! makes sense.

Cheers.

Couple prelim questions (1 Viewer)

piethepker

Active Member

InteGrand

Well-Known Member

piethepker

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)