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cupoosta [38]
3 years ago
14

You are provided with a compound fertilizer, 40-15-10. Calculate the quantity of fertilizer to add to a one hectare field to sup

ply a. Nitrogen at 120 kg/ha (2 marks) b. Nitrogen at 90 kg/ha (2 marks) c. Phosphorus (P 2 O 5 ) at rate of 60 kg/ha (2 marks) d. Potassium (K 2 O) at rate of 60 kg/ha
Chemistry
1 answer:
Kryger [21]3 years ago
5 0

Answer:

The correct answer is a) 300 Kg b) 225 Kg c) 400 Kg and d) 600 Kg.

Explanation:

Based on the given value, that is. 40-15-10 shows the compositions of nitrogen, phosphorus, and potassium found in the compound fertilizer in the form of percentage.  

41 percent nitrogen is equal to 40/100 = 0.4, 15 percent phosphorus is equal to 15/100 = 0.15, and 10 percent potassium is equal to 10/100 = 0.1.  

To find the amount of fertilizer, which is needed per hectare to supply the different nutrients per hectare, there is a need to divide the quantity to be supplied by the percentage composition, now the values comes out as:  

a) At 120 Kg per ha, the required nitrogen will be 120/0.4 = 300 Kilogram of fertilizer.  

b) At 90 kg per ha, the required nitrogen will be 90/0.4 = 225 Kilograms of fertilizer.  

c) At 60 kg per ha, the required phosphorus will be 60/0.15 = 400 Kilograms of fertilizer.  

d) At 60 Kg per ha, the required potassium will be 60/0.1 = 600 Kilograms of fertilizer.  

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\text{Molarity}=\frac{\text{Moles}}{\text{Volume}} ......(1)

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Molarity of iron (II) sulfate = 1 M

Volume of solution = 200 mL = 0.200 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

\text{Moles of }FeSO_4=(1mol/L\times 0.200L)=0.200mol

The chemical equation for the reaction of FeO with sulfuric acid follows:

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By stoichiometry of the reaction:

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Putting values in above equation, we get:

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Hence, the mass of FeO required is 14.37 g

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