5.1) A well-known ratio of chemicals Nitrogen ( N\( ) \), Phosphorus \( (\mathrm{P}) \) and Potassium \( (\mathrm{K}) \) in the form a fertilizer, in the ratio \( 3: 2: 1 \), are put in a bag of 1250 g as a mixture. Determine the grams of \( \mathrm{N}, \mathrm{P} \) and K that make up the mixture of 1250 g . (7)

To determine the grams of Nitrogen (N), Phosphorus (P), and Potassium (K) in a 1250 g mixture with a ratio of 3:2:1, we can follow these steps: 1. **Identify the total parts in the ratio**: The ratio of N:P:K is 3:2:1. To find the total number of parts, we add the parts together: \[ \text{Total parts} = 3 + 2 + 1 = 6 \] 2. **Calculate the weight of each component**: - The weight of Nitrogen (N) can be calculated as: \[ N = \left(\frac{3}{6}\right) \times 1250 \] - The weight of Phosphorus (P) can be calculated as: \[ P = \left(\frac{2}{6}\right) \times 1250 \] - The weight of Potassium (K) can be calculated as: \[ K = \left(\frac{1}{6}\right) \times 1250 \] Now, let's perform the calculations for each component. ### Calculations 1. **Calculate N**: \[ N = \left(\frac{3}{6}\right) \times 1250 = \frac{3 \times 1250}{6} = \frac{3750}{6} = 625 \text{ g} \] 2. **Calculate P**: \[ P = \left(\frac{2}{6}\right) \times 1250 = \frac{2 \times 1250}{6} = \frac{2500}{6} \approx 416.67 \text{ g} \] 3. **Calculate K**: \[ K = \left(\frac{1}{6}\right) \times 1250 = \frac{1 \times 1250}{6} = \frac{1250}{6} \approx 208.33 \text{ g} \] ### Summary of Results - Grams of Nitrogen (N): \( 625 \text{ g} \) - Grams of Phosphorus (P): \( 416.67 \text{ g} \) - Grams of Potassium (K): \( 208.33 \text{ g} \) Thus, the mixture of 1250 g contains approximately: - \( 625 \text{ g} \) of Nitrogen, - \( 416.67 \text{ g} \) of Phosphorus, - \( 208.33 \text{ g} \) of Potassium.