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A multi-nutrient fertilizer contains several different nitrogen containing compounds. The fertilizer is 52.2% CH4N2O (urea), 22.0% KNO₃ , and 12.0% (NH₄)2HPO₄ by mass. The remainder of the fertilizer consists of substances that do not contain nitrogen. How much fertilizer should someone apply to provide 2.50 g N to a plant?

User Bhaller
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Final answer:

To supply 2.50 g of nitrogen to a plant, one must apply 8.38 g of the multi-nutrient fertilizer. This calculation involves determining the mass percentage of nitrogen in each component of the fertilizer and using it to calculate the overall nitrogen content.

Step-by-step explanation:

To determine how much fertilizer is needed to provide 2.50 g of nitrogen to a plant, we must consider the percentage of nitrogen in each component of the fertilizer. The three nitrogen-containing compounds in the multi-nutrient fertilizer are urea (CH4N2O), potassium nitrate (KNO₃), and diammonium phosphate ((NH₄)2HPO₄).

Firstly, we calculate the mass percentage of nitrogen in each compound:

  • Urea (CH4N2O) contains 46.65% nitrogen by mass.
  • Potassium nitrate (KNO₃) contains 13.85% nitrogen by mass.
  • Diammonium phosphate ((NH₄)2HPO₄) contains 21.21% nitrogen by mass.

Next, we use these percentages along with the mass percentages of each compound in the fertilizer to calculate the overall mass percentage of nitrogen in the fertilizer:

Total nitrogen percentage in fertilizer = (52.2% × 46.65%) + (22.0% × 13.85%) + (12.0% × 21.21%) = 24.2433% + 3.047% + 2.5452% = 29.8355%

Now, to find how much fertilizer is needed to provide 2.50 g of nitrogen, we use the proportion:

× (amount of fertilizer) × 29.8355% = 2.50 g of N

× = 2.50 g / 0.298355 = 8.38 g of fertilizer

Therefore, to supply 2.50 g of nitrogen, one would need to apply 8.38 g of the multi-nutrient fertilizer.

User GoldenJam
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