This is a sequential organic synthesis problem where we track the moles of the limiting reactant (Benzene) through three steps, calculating the yield and molecular weight changes at each stage.

1. Determine Initial Moles of Benzene

Benzene formula: $\text{C}_6\text{H}_6$.

Molar Masses (g/mol): $\text{H}=1, \text{C}=12$.

$$MW_{\text{Benzene}} = 6(12) + 6(1) = 78 \text{ g/mol}$$

Initial mass of Benzene: $7.8 \text{ g}$.

$$\text{Moles of Benzene} = \frac{7.8 \text{ g}}{78 \text{ g/mol}} = 0.10 \text{ mol}$$

2. Reaction Sequence Analysis

Step 1: Benzene $\xrightarrow{\text{oleum (excess), } 200^\circ\text{C}} \text{P}$ (Sulfonation)

At high temperature ($200^\circ\text{C}$), sulfonation of Benzene yields 1,3-Benzenedisulfonic acid ($\text{C}_6\text{H}_4(\text{SO}_3\text{H})_2$). The reaction is assumed to be substitution of two $\text{H}$ by two $\text{SO}_3\text{H}$ groups.

$$MW_{\text{P}} = MW_{\text{Benzene}} - 2\cdot 1 + 2 \cdot (\text{S} + 3\text{O} + \text{H})$$

$$MW_{\text{P}} = 78 - 2 + 2(32 + 3\cdot 16 + 1) = 76 + 2(81) = 76 + 162 = 238 \text{ g/mol}$$

Yield: $80\%$.

$$\text{Moles of P} = 0.10 \text{ mol} \times 0.80 = 0.080 \text{ mol}$$

Step 2: $\text{P} \xrightarrow{\text{NaOH, heat then } \text{H}_3\text{O}^+} \text{Q}$ (Base Fusion/Hydroxylation)

Benzenedisulfonic acid ($\text{P}$) reacts with $\text{NaOH}$ fusion to replace the sulfonate groups ($\text{SO}_3\text{H}$) with hydroxyl groups ($\text{OH}$), followed by acidification, yielding Resorcinol (1,3-Dihydroxybenzene, $\text{C}_6\text{H}_4(\text{OH})_2$).

$$MW_{\text{Q}} = MW_{\text{Benzene}} - 2 \cdot 1 + 2 \cdot (\text{O} + \text{H})$$

$$MW_{\text{Q}} = 78 - 2 + 2(16 + 1) = 76 + 34 = 110 \text{ g/mol}$$

Yield: $75\%$.

$$\text{Moles of Q} = 0.080 \text{ mol} \times 0.75 = 0.060 \text{ mol}$$

Step 3: $\text{Q} \xrightarrow{\text{HNO}_3 \text{ (excess)} / \text{H}_2\text{SO}_4 \text{ (excess)}} \text{R}$ (Nitration)

Resorcinol ($\text{Q}$) is highly activated. Excess nitrating mixture substitutes all available positions (2, 4, 6 positions relative to $\text{OH}$ groups are active). Assuming complete nitration of all free positions (2, 4, 6, 5), but based on standard chemistry, Resorcinol yields 2,4,6-trinitroresorcinol (Styphnic acid, $\text{C}_6(\text{NO}_2)_3(\text{OH})_2\text{H}$). Wait, only three positions are substituted (2, 4, 6 positions relative to the two $\text{OH}$ groups). The final product $\text{R}$ is 2,4,6-trinitroresorcinol.

Formula of $\text{R}$ ($\text{C}_6(\text{OH})_2\text{H}_3(\text{NO}_2)_3$): $\text{C}_6\text{H}_3\text{O}_2(\text{NO}_2)_3$.

Change: $3 \text{ H}$ atoms replaced by $3 \text{ NO}_2$ groups.

$$MW_{\text{R}} = MW_{\text{Q}} - 3\cdot 1 + 3 \cdot (\text{N} + 2\text{O})$$ $$MW_{\text{R}} = 110 - 3 + 3 \cdot (14 + 32) = 107 + 3 \cdot 46$$ $$MW_{\text{R}} = 107 + 138 = 245 \text{ g/mol}$$

Yield: $50\%$.

$$\text{Moles of R} = 0.060 \text{ mol} \times 0.50 = 0.030 \text{ mol}$$

3. Calculate Final Mass of R

$$\text{Mass of R} = \text{Moles of R} \times MW_{\text{R}}$$ $$\text{Mass of R} = 0.030 \text{ mol} \times 245 \text{ g/mol}$$ $$\text{Mass of R} = 7.35 \text{ g}$$

4. Final Formatting

The amount of $\text{R}$ produced is $7.35 \text{ g}$. Rounding off to two decimal places gives 7.35.

This matches the provided constraint range [7.35, 7.35].