Final answer:
The maximal amount of HClO₃ producible from 7.9 grams of Cl₂ is calculated through stoichiometric calculations involving the conversion of mass to moles of Cl₂, using the stoichiometric ratio from the balanced equation to find moles of HClO₃, and then converting back to grams.
Step-by-step explanation:
When you start with 7.9 grams of Cl₂ and it is the limiting reactant, the maximum amount of HClO₃ you can produce can be determined through stoichiometric calculations. These calculations are based on the chemical equation of the reaction, which must be known or provided to solve the problem. Since the specific chemical reaction to produce HClO₃ is not given in your question, a general approach can be outlined.
To solve such problems, one should first convert the mass of the limiting reactant to moles, using its molar mass. Then use the stoichiometry of the balanced chemical equation, to find the molar ratio between the limiting reactant and the desired product. Finally, convert moles of product to grams using the molar mass of the product.
If the balanced equation for the formation of HClO₃ from Cl₂ was provided, you would follow these steps, substituting in the appropriate molar ratios. Without the actual reaction, an exact numerical answer cannot be given, but this process defines the method you would use to determine the maximal amount of HClO₃ producible from 7.9 grams of Cl₂.