In the following sequence of reactions, the amount of D (in g) formed from 72 g of Mg is……….. (The yield (%) corresponding to the product in each step is given in the parenthesis)
`underset(72g)Mgunderset("Strong heating")overset(N_(2),"excess")tounderset(100%)Aoverset(H_(2)O)tounderset((90%))Boverset("conc."HNO_(3))tounderset((80%))Coverset("Heat")tounderset((90%))D`

Explanation

To determine the amount of product D formed from 72 g of magnesium (Mg) through a series of reactions, we will follow the steps outlined in the question and apply the given yield percentages at each stage.

### Step-by-Step Solution:

1. **Identify the Reaction of Magnesium with Nitrogen:**
- The first reaction involves magnesium reacting with nitrogen (N₂) to form magnesium nitride (Mg₃N₂).
- The balanced equation for this reaction is:
\[ 3 \text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3\text{N}_2 \]
- Molar mass of Mg = 24 g/mol. Therefore, 72 g of Mg corresponds to:
\[ \text{Moles of Mg} = \frac{72 \text{ g}}{24 \text{ g/mol}} = 3 \text{ moles} \]
- From the balanced equation, 3 moles of Mg produce 1 mole of Mg₃N₂. Thus, 3 moles of Mg will produce:
\[ 1 \text{ mole of Mg}_3\text{N}_2 \]

2. **Hydrolysis of Magnesium Nitride:**
- The next step is the hydrolysis of Mg₃N₂ to form ammonia (NH₃):
\[ \text{Mg}_3\text{N}_2 + 6 \text{H}_2\text{O} \rightarrow 3 \text{NH}_3 + 3 \text{Mg(OH)}_2 \]
- From 1 mole of Mg₃N₂, we get 3 moles of NH₃. Therefore, from 1 mole of Mg₃N₂, we will produce:
\[ 3 \text{ moles of NH}_3 \]

3. **Reaction of Ammonia with Concentrated Nitric Acid:**
- The ammonia then reacts with concentrated nitric acid (HNO₃) to form ammonium nitrate (NH₄NO₃):
\[ \text{NH}_3 + \text{HNO}_3 \rightarrow \text{NH}_4\text{NO}_3 \]
- The yield for this reaction is 80%. Thus, the amount of NH₄NO₃ produced from 3 moles of NH₃ is:
\[ \text{Moles of NH}_4\text{NO}_3 = 3 \text{ moles} \times 0.80 = 2.4 \text{ moles} \]

4. **Heating Ammonium Nitrate:**
- Heating NH₄NO₃ produces nitrous oxide (N₂O):
\[ 2 \text{NH}_4\text{NO}_3 \rightarrow 2 \text{N}_2\text{O} + 2 \text{H}_2\text{O} + \text{N}_2 \]
- From 2 moles of NH₄NO₃, we get 2 moles of N₂O. Therefore, from 2.4 moles of NH₄NO₃, we will produce:
\[ \text{Moles of N}_2\text{O} = 2.4 \text{ moles} \times 1 = 2.4 \text{ moles} \]

5. **Calculating the Mass of N₂O:**
- Molar mass of N₂O = (2 × 14) + 16 = 44 g/mol.
- Therefore, the mass of N₂O produced is:
\[ \text{Mass of N}_2\text{O} = 2.4 \text{ moles} \times 44 \text{ g/mol} = 105.6 \text{ g} \]

6. **Final Yield Calculation:**
- The final yield of N₂O is 90%. Thus, the actual amount of D (N₂O) produced is:
\[ \text{Final mass of D} = 105.6 \text{ g} \times 0.90 = 95.04 \text{ g} \]

### Conclusion:
The amount of D formed from 72 g of magnesium is approximately **95.04 g**.