Problem 1

Diatoms, microscopic organisms, produce carbohydrates from carbon dioxide and water by photosynthesis:

\ce{6CO2 + 6H2O ->[\text{solar energy}] C6H12O6 + 6O2}

. (1.1) During their first five years blue whales gain 75 kg of mass per day by feeding on krill; the whale must consume ten times this mass of krill each day, and the krill must consume 10.0 kg of diatoms to produce 1.0 kg of krill. Assuming the whale's mass gain comes from carbohydrate consumption, calculate the volume of

\ce{CO2}

at STP (0 °C, 101 kPa) the diatoms must fix to feed one whale for its first five years. (1.2) Sea water contains

0.23\ \mathrm{cm^3}

of dissolved

\ce{CO2}

per litre (24 °C, 101 kPa). (i) What volume of water must be processed to supply this

\ce{CO2}

? (ii) What fraction of the ocean volume (

1.37\times10^{18}\ \mathrm{m^3}

) is needed for 1000 whales? (1.3) Three percent of a

9.1\times10^{4}

kg adult whale is nitrogen; what maximum mass of

\ce{NH4+}

can a dead whale release? (1.4) Eighteen percent of the whale is carbon, returned to the atmosphere as

\ce{CO2}

;

\ce{CO2}

is removed by weathering

\ce{CaSiO3(s) + 2CO2(g) + 3H2O(l) -> Ca^{2+}(aq) + 2HCO3^-(aq) + H4SiO4(aq)}

. What mass of

\ce{CaSiO3}

can be weathered by the

\ce{CO2}

from 1000 dead whales?

Step 2 of 4: CO₂ volume at STP

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n(\ce{CO2}) = \frac{1.4\times10^{7}\ \mathrm{kg}}{0.180\ \mathrm{kg\,mol^{-1}}} \times 6 = 4.5\times10^{8}\ \mathrm{mol};\qquad V = 4.5\times10^{8} \times 22.4\ \mathrm{dm^3} \approx 1.0\times10^{10}\ \mathrm{dm^3}

Analysis

Six moles of $\ce{CO2}$ make one mole of $\ce{C6H12O6}$ (180 g), equivalent to $1.4\times10^{7}\times264/180 = 2.0\times10^{7}$ kg $\ce{CO2}$ $= 4.5\times10^{8}$ mol; at 22.4 dm³ mol⁻¹ this is $1.0\times10^{10}$ dm³.