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nutrients in plant fertilizers. By industry convention, the numbers on a label refer to the mass percents of $\mathrm{N}, \mathrm{P}_{2} \mathrm{O}_{5},$ and $\mathrm{K}_{2} \mathrm{O}$ , in that order.

Calculate the N/P/K ratio of a 30$/ 10 / 10$ fertilizer in terms of moles of each element, and express it as $x / y / 1.0 .$

$\mathrm{N} / \mathrm{P} / \mathrm{K}=7 / 1 / 2$ or to be more precise $\mathrm{N} / \mathrm{P} / \mathrm{K}=6,9 / 1 / 1,9$

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Numerade Educator

University of Kentucky

Numerade Educator

Brown University

Gamble fertilizer that has a ratio of the components and that we can find this ratio in moles, this ratio that's given to us by Mass. So we need to change it to moles, and we do that using the molar mass of each substance. So we'll need to know. The molar mass of nitrogen, which we find on the periodic table, is equal to 14 wait 007 grams per mole. The Mueller, Mass. Of P 205 is found by taking the more massive phosphorus, multiplying it by two and adding five times the molar mass of oxygen. And this is 100 and 41 0.94 grams from all. And finally, the molar mass of potassium oxide, which is found similarly two times. Potassium plus one oxygen or 94 went 195 and we're looking for the ratio of nitrogen phosphorus to potassium. So that's the second thing we'll have to think about in terms of taking into account that these are compounds and these are elements since it's by Mass, if it's 30%. If we were to have 50 grams, that means that 30 grams of it would be nitrogen, which we can change two moles by dividing by the molar mass, which equals 2.14 moles. Nitrogen. Similarly, we would have 10 grams of P 205 which we changed to moles by dividing by the molar mass. But then we need to find moles of phosphorus. And so according to the formula, there are two moles of p for every one mole of compound and so this equals 0.141 moles a class for us. Similarly, we confined the moles of potassium given that if we had 50 grams, there would be 10 grams of potassium oxide to change two moles. And again, there are two moles of the element for every one mole of the compound or 0.212 moles. Okay, we can write this as the ratio and two p two k equals 2.14 0.1 for one to 0.212 and then because we know we want the K two b one. We divide each of these numbers by the percentage of K and moles, So divide each number I 0.212 which gives us a ratio of 10 2.66 to one sequel to our ratio