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).
Is there a question on this poroblems? are you grazing for answers? We will be happy to critique your work.
This is the same person who posted six or more in a big HOMEWORK DUMP yesterday, too. No thinking involved. Just posts and hopes someone will do his work for him. =(
a) To estimate the increase in the amount of recyclable P in a country of 5.5 million persons in one year, we need to calculate the difference in the amount of P recycled between the conventional flushing toilets and the alternative dry toilets.
In the conventional flushing toilet system, only 1% of the excreted P is recycled into food production. Therefore, the amount of P recycled per person per day is:
1% of 2.2 g = 0.01 * 2.2 g = 0.022 g
In the alternative dry toilet system, 80% of the excreted P is recycled into food production. Therefore, the amount of P recycled per person per day is:
80% of 2.2 g = 0.8 * 2.2 g = 1.76 g
Now, let's calculate the total amount of P recycled in one year for a country of 5.5 million persons:
Total P recycled in one day = 1.76 g * 5.5 million = 9.68 million grams or 9680 metric tons.
Therefore, the increase in the amount of recyclable P in a country of 5.5 million persons in one year would be approximately 9680 metric tons.
b) To calculate the mass percentages of P and N in struvite, (NH4MgPO4), we need to determine the molar mass of each element and calculate their respective mass percentages.
The molar masses of the elements are:
P (phosphorus) = 30.97 g/mol
N (nitrogen) = 14.01 g/mol
The molar mass of the struvite compound, (NH4MgPO4), can be calculated as follows:
Molar mass of (NH4MgPO4) = (1 * molar mass of N) + (4 * molar mass of H) + (1 * molar mass of Mg) + (1 * molar mass of P) + (4 * molar mass of O)
Molar mass of (NH4MgPO4) = (1 * 14.01 g/mol) + (4 * 1.01 g/mol) + (1 * 24.31 g/mol) + (1 * 30.97 g/mol) + (4 * 16.00 g/mol)
Molar mass of (NH4MgPO4) = 132.36 g/mol
Now, let's calculate the mass percentages of P and N in struvite:
Mass percentage of P = (molar mass of P / molar mass of (NH4MgPO4)) * 100%
Mass percentage of P = (30.97 g/mol / 132.36 g/mol) * 100%
Mass percentage of P ≈ 23.43%
Mass percentage of N = (molar mass of N / molar mass of (NH4MgPO4)) * 100%
Mass percentage of N = (14.01 g/mol / 132.36 g/mol) * 100%
Mass percentage of N ≈ 10.58%
Therefore, the mass percentages of P and N in struvite, (NH4MgPO4), are approximately 23.43% and 10.58%, respectively.