Question

At the end of 2012, global population was about 7.0 billion people. What mass of glucose in kg would be needed to provide 1900 cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to CO2(g) and H2O(l) according to the following thermochemical equation: C6H12O6(s)+6O2(g)⟶6CO2(g)+6H2O(l)ΔH∘=−2803kJ

Answer 1

daily need per person = 1900 Cal/person/day

number of days = 1 year = 365 days

total person = 7.0 billion = 7.0 x 10⁹

Total energy required = (daily need per person) * (number of days) * (total person)

Total energy required = (1900 Cal/person/day) * (365 days) * (7.0 x 10⁹ person)

Total energy required = 4.8545 x 10¹⁵ Cal

Total energy required = 4.8545 x 10¹⁵ Cal * (4.184 kJ / Cal)

Total energy required = 2.03 x 10¹⁶ kJ

moles of glucose = (total energy) / (enthalpy of reaction)

moles of glucose = (2.03 x 10¹⁶ kJ) / (2803 kJ/mol)

moles of glucose = 7.246 x 10¹² mol

mass of glucose = (moles of glucose) * (molar mass C₆H₁₂O₆)

mass of glucose = (7.246 x 10¹² mol) * (180.156 g/mol)

mass of glucose = 1.3 x 10¹⁵ g

mass of glucose = 1.3 x 10¹⁵ g * (1 kg / 1000 g)

mass of glucose = 1.3 x 10¹² kg