A homeowner is frustrated with being the laughing-stock of the neighborhood beca

Question

A homeowner is frustrated with being the laughing-stock of the neighborhood because his yard is in awful shape. He has a conversation with your and you recommend he has a soil test performed on his lawn. Instead of going through the trouble himself, he offers to pay you to get his yard back into shape. The soil test recommends you apply 2 lbs of N/1000 ft' twice during the summer and 0.75 ft of P/1000ft' on the 1 of July. The lot size is 300 times 250' and the house and landscape take up 7500 square feet. How many square feet of turf must be fertilized? How much fertilizer are you going to put out for ONE of the Nitrogen applications if you use area fertilizer [46-0-0]? How much will you apply for the phosphorus application using TSP [0-32-0]?

Explanation / Answer

SOLUTION:

(a). Lot size = 300 X 250 = 75000 sq ft

space occupied by house and landscape = 7500 sq ft

Hence size of turf to be fertilized = 75000 - 7500 = 67500 sq ft

(b) Nitrogen need per 1000 ft = 2lb (907.2g)

Nitrogen needed for 67500 sq ft = (2/1000) X 67500 = 135 lb = 61235g

Molar mass of urea (CH4N2O) = 60g

It means 60g of urea gives 28g of nitrogen, because two moles of nitrogen atom are present in 60g of urea.

28g of Nitrogen = 60g f urea

1g of nitrogen will be had from = 60/28 = 2.14g of urea.

Therefore 61235g of of nitrogen will be available from = 61235 X 2.14 = 131,218g of urea = 289.2 lb of urea.

So one fertilizer application amounts to 289.2lb of the fertilizer.

(c). Phosphorus needed for 67500 sq ft = (0.75/1000) X 67500 = 50.6 lb = 22952g

TSP contains 32% P (given). It implies 32g of P will be had from 100g of TSP

22952g of P will be available from = (100/32) X 22952 = 819714g = 1807.2 lb of TSP.

Hence we need to apply 1807 lb of Fertiliser.

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