A fertilizer contains 55.1% CH,N,O (urea), 26.8% KNO, and 14.8% (NH),HPO, by mass.How much fertilizer should someone apply to provide 2.00 g N to a plant?

Answer 1

You might want to go back and re-submit the question with the correct chemical formulas. I'm guessing you meant to write

55.1% CH₄N₂O (urea)

26.8% KNO₃

14.8% (NH₄)₂HPO₄

The way you have the compounds written is not possible as these do not exist. I would go ahead and answer the question with my "guessed" compounds, but if that's not what you meant, it would simply be a waste of time. Please be careful when copying and submitting your questions. Thanks.

Answer 2

1. Calculate the mass of nitrogen in each component based on its percentage:

Urea (CH₄N₂O):

Mass percentage of N in urea = 2(N) / (12(C) + 4(H) + 2(N) + 16(O)) * 100% = (2/60) * 100% = 3.33%

Potassium nitrate (KNO₃):

Mass percentage of N in KNO3 = 1(N) / (39.1(K) + 14(N) + 3(16(O)) * 100% = (1/101.1) * 100% = 0.99%

Ammonium dihydrogen phosphate ((NH₄)H₂PO₄):

Mass percentage of N in (NH₄)H₂PO₄ = 2(N) / [(2(14(N)) + 8(H) + 16(O)) + (31(P) + 16(O) + 4(H))] * 100% = (2/115.1) * 100% = 1.74%

Calculate the total mass of the fertilizer required to provide 2.00 grams of nitrogen:

Let F be the mass of fertilizer needed (in grams).

The total mass of nitrogen from each component is as follows:

From urea: 3.33% of F grams

From potassium nitrate: 0.99% of F grams

From ammonium dihydrogen phosphate: 1.74% of F grams

The total mass of nitrogen should equal 2.00 grams:

0.0333F+0.0099F+0.0174F=2.00

Now, solve for F:

0.0333F+0.0099F+0.0174F=2.00

0.0606F=2.00

F= 2.00/0.0606

≈33.00 grams

So, someone should apply approximately 33.00 grams of the fertilizer to provide 2.00 grams of nitrogen to a plant.