Question

Nitrogen (N), phosporus (P) and potassium (K) are the main nutrients in plant fertilizers. . According to an industry convention , the numbers on the label refer to the mas % of N, P 2 O 5 and K 2 O in that order. If the N:P:K (in terms of moles of each elements) ratio of a fertilizer labelled as 28:11.75:11.75fertilizer in terms of moles of each elements, is expressed as x:y:1, then find the value of y.

Answer 1

0.66

Answer 2

To determine the N:P:K molar ratio, we first need to understand the industry convention for fertilizer labels. The numbers represent the mass percentage of N, P₂O₅, and K₂O, respectively.

Given fertilizer label: 28:11.75:11.75

This implies:

• Mass % of N = 28%
• Mass % of P₂O₅ = 11.75%
• Mass % of K₂O = 11.75%

Let's assume we have 100 g of the fertilizer. Then, the mass of each component is:

• Mass of N = 28 g
• Mass of P₂O₅ = 11.75 g
• Mass of K₂O = 11.75 g

Now, we need to convert these masses into moles of the respective elements (N, P, K). We use the approximate molar masses:

• Molar mass of N = 14 g/mol
• Molar mass of P = 31 g/mol
• Molar mass of K = 39 g/mol
• Molar mass of O = 16 g/mol

1. Moles of N: The label directly gives the mass % of N. Moles of N = (Mass of N) / (Molar mass of N) Moles of N = 28 g / 14 g/mol = 2 mol

2. Moles of P: The label gives the mass % of P₂O₅. We need to find moles of P from P₂O₅. Molar mass of P₂O₅ = (2 × Molar mass of P) + (5 × Molar mass of O) Molar mass of P₂O₅ = (2 × 31) + (5 × 16) = 62 + 80 = 142 g/mol Moles of P₂O₅ = (Mass of P₂O₅) / (Molar mass of P₂O₅) Moles of P₂O₅ = 11.75 g / 142 g/mol Since 1 mole of P₂O₅ contains 2 moles of P, Moles of P = 2 × (Moles of P₂O₅) = 2 × (11.75 / 142) mol

3. Moles of K: The label gives the mass % of K₂O. We need to find moles of K from K₂O. Molar mass of K₂O = (2 × Molar mass of K) + (1 × Molar mass of O) Molar mass of K₂O = (2 × 39) + (1 × 16) = 78 + 16 = 94 g/mol Moles of K₂O = (Mass of K₂O) / (Molar mass of K₂O) Moles of K₂O = 11.75 g / 94 g/mol Since 1 mole of K₂O contains 2 moles of K, Moles of K = 2 × (Moles of K₂O) = 2 × (11.75 / 94) mol

Now, we have the molar ratio N:P:K:

N : P : K = 2 : [2 × (11.75 / 142)] : [2 × (11.75 / 94)]

We need to express this ratio as x:y:1. To do this, we divide all terms by the moles of K:

x = Moles of N / Moles of K y = Moles of P / Moles of K 1 = Moles of K / Moles of K

Let's calculate y:

$y = \frac{2 \times \frac{11.75}{142}}{2 \times \frac{11.75}{94}}$

The terms '2' and '11.75' cancel out:

$y = \frac{\frac{1}{142}}{\frac{1}{94}}$

$y = \frac{94}{142}$

$y \approx 0.66197$

Rounding to two decimal places, y ≈ 0.66.

For completeness, let's also calculate x:

$x = \frac{2}{2 \times \frac{11.75}{94}}$

$x = \frac{1}{\frac{11.75}{94}}$

$x = \frac{94}{11.75}$

$x = 8$

So the N:P:K ratio is 8:0.66:1. The value of y is 0.66.