Question

In order to carry out an anaerobic fermentation, a saturated solution of strontium hydroxide is used in a trap. The strontium hydroxide reacts with the carbon dioxide produced from the fermentation yielding a precipitate of strontium carbonate, while no air is allowed into the system. How many grams of strontium carbonate would you expect to collect from the fermentation of 55.00g of sucrose at 65.0 % efficiency and at 23 C and 1 atm? Assume all CO2 is trapped as strontium carbonate.

Please explain how you know what the chemical equation should be.

Answer 1

The chemical equation for the fermentation of sucrose, followed by production of SrCO₃ is shown below.

i.e. The fermentation of 1 mole of sucrose at 100% efficiency causes the formation of 4 moles of SrCO₃.

Therefore, the fermentation of 1 mole of sucrose at 65% efficiency causes the formation of (65/100)*4, i.e. 2.6 moles of SrCO₃.

The mass of sucrose undergoing fermentation = 55 g

The molar mass of sucrose = 342.297 g/mol

The no. of moles of sucrose undergoing fermentation = 55/342.297, i.e. 0.161

Therefore, the fermentation of 0.161 moles of sucrose at 65% efficiency causes the formation of 0.161*2.6 = 0.418 moles of SrCO₃.

The molar mass of SrCO₃ = 147.628

Therefore, the amount of strontium carbonate = 0.418*147.628, i.e. 61.71 g.