10.1 The flow diagram below shows how fertiliser B is produced in industry - NSC Physical Sciences - Question 10 - 2020 - Paper 2

T is known as Natural gas.

The catalyst used in process 1 is Iron (Fe) or Iron oxide (FeO).

Compound A is Ammonia (NH₃).

Process 2 is known as the Ostwald process.

The balanced equation for the formation of fertiliser B is:

$NH₃ + HNO₃ \rightarrow NH₄NO₃$

The ratio represents the proportion of the three primary nutrients in the fertiliser: Nitrogen, Phosphorus, and Potassium.

To calculate X, we first determine the total parts in the ratio: 2 + 4 + 3 = 9 parts.

Now, 20 kg corresponds to these 9 parts, so each part is:

$\text{Mass of each part} = \frac{20 \text{ kg}}{9} \approx 2.222 \text{ kg}$

Thus, the mass of each component will be:

• Nitrogen = 2 parts = $2 \times 2.222 \approx 4.44$ kg
• Phosphorus = 4 parts = $4 \times 2.222 \approx 8.888$ kg
• Potassium = 3 parts = $3 \times 2.222 \approx 6.666$ kg

Now, using the information that the bag contains 2,315 kg of phosphorous:

$4X = 2315 \text{ kg} \Rightarrow X = \frac{2315}{4} \approx 578.75\, ext{ kg}$