Phosphorous (P) is an indispensable nutrient in food production. According to the estimates, mineral
phosphorous resources will end up within 50 years and consequently new sources of P has to be found.
A human excretes 2,2 g of phosphorous (P) per day on average. In conventional wastewater management
system, with flushing toilets, only 1 % of the excreted P is recycled into food production. Dry toilets
could be used as an alternative to the conventional flushing toilets and they can make it possible to recycle
80 % of P directly into food production.
a) Estimate the increase in amount of recyclable P in a country of 5,5 million persons in one year, if alternative
dry toilets would replace the conventional flushing toilets.
b) Calculate the mass percentages of P and N in struvite, (NH4MgPO4).
a) To estimate the increase in the amount of recyclable P in a country of 5.5 million persons in one year, we first need to calculate the amount of P excreted by one person in one year.
Given that a human excretes 2.2 g of phosphorous (P) per day on average, we can calculate the annual amount:
2.2 g/day * 365 days/year = 803 g/year
Next, we need to calculate the difference in the recycling percentage between conventional flushing toilets (1% recycled) and alternative dry toilets (80% recycled).
For conventional flushing toilets, only 1% of the excreted P is recycled, which means:
1% * 803 g/year = 8.03 g/year recycled per person
For alternative dry toilets, 80% of the excreted P is recycled, which means:
80% * 803 g/year = 642.4 g/year recycled per person
Now, to estimate the increase in the amount of recyclable P in a country of 5.5 million persons in one year, we can multiply the recycled amount per person by the population:
For conventional flushing toilets:
8.03 g/year/person * 5.5 million persons = 44,165,000 g/year or 44,165 kg/year
For alternative dry toilets:
642.4 g/year/person * 5.5 million persons = 3,531,200,000 g/year or 3,531,200 kg/year
Therefore, by switching to alternative dry toilets, the amount of recyclable phosphorous (P) would increase by approximately 3,486,035 kg/year.
b) To calculate the mass percentages of P and N in struvite, (NH4MgPO4), we need to determine the molar mass of each element and divide it by the molar mass of the compound.
The molar mass of P is 30.97 g/mol and the molar mass of N is 14.01 g/mol.
The molar mass of (NH4MgPO4) can be calculated as follows:
(1 * molar mass of N) + (4 * molar mass of H) + molar mass of Mg + molar mass of P + (4 * molar mass of O)
(1 * 14.01 g/mol) + (4 * 1.01 g/mol) + 24.31 g/mol + 30.97 g/mol + (4 * 16.00 g/mol)
14.01 g/mol + 4.04 g/mol + 24.31 g/mol + 30.97 g/mol + 64.00 g/mol
137.33 g/mol
Now, let's calculate the mass percentage of each element in (NH4MgPO4):
Mass percentage of P:
(molar mass of P / molar mass of (NH4MgPO4)) * 100
(30.97 g/mol / 137.33 g/mol) * 100
22.56%
Mass percentage of N:
(molar mass of N / molar mass of (NH4MgPO4)) * 100
(14.01 g/mol / 137.33 g/mol) * 100
10.19%
Therefore, the mass percentages of P and N in struvite, (NH4MgPO4), are approximately 22.56% and 10.19%, respectively.