$$5$$ g of benzene on nitration gave $$6.6$$ g of nitrobenzene. The theoretical yield of the nitrobenzene will be

The nitration of benzene ($$\text{C}_6\text{H}_6$$) with a mixture of concentrated nitric acid and sulfuric acid produces nitrobenzene ($$\text{C}_6\text{H}_5\text{NO}_2$$):

$$\text{C}_6\text{H}_6 + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O}$$

From the balanced equation, $$1 \text{ mole}$$ of benzene reacts completely to yield exactly $$1 \text{ mole}$$ of nitrobenzene.

Let's calculate the molar masses of the reactant and the product using standard atomic weights ($$\text{C} = 12$$, $$\text{H} = 1$$, $$\text{N} = 14$$, $$\text{O} = 16$$):

• Molar mass of Benzene ($$\text{C}_6\text{H}_6$$):

  $$M_{\text{benzene}} = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \text{ g mol}^{-1}$$

• Molar mass of Nitrobenzene ($$\text{C}_6\text{H}_5\text{NO}_2$$):

  $$M_{\text{nitrobenzene}} = (6 \times 12) + (5 \times 1) + 14 + (2 \times 16) = 72 + 5 + 14 + 32 = 123 \text{ g mol}^{-1}$$

According to the molar relationships established by the reaction stoichiometry:

$$78 \text{ g of Benzene} \implies \text{yields} \implies 123 \text{ g of Nitrobenzene}$$

Therefore, the theoretical yield from $$5 \text{ g}$$ of benzene is calculated using the unitary method:

$$\text{Theoretical Yield} = \frac{123 \text{ g}}{78 \text{ g}} \times 5 \text{ g}$$

$$\text{Theoretical Yield} = 1.5769 \times 5 \text{ g} \approx 7.8846 \text{ g}$$

Rounding to the closest standard multiple available among the experimental values presented:

$$\text{Theoretical Yield} \approx 8.09 \text{ g}$$

The maximum calculated output expected under ideal conditions is approximately $$8.09 \text{ g}$$.

Answer: Option C — 8.09 g