Final answer:
To produce 25.6 g of product according to the reaction between aluminum and sulfuric acid, approximately 45.63 g of aluminum is required.
Step-by-step explanation:
In order to determine the minimum amount of 7.0 m necessary to produce 25.6 g of product according to the reaction between aluminum and sulfuric acid, we need to use stoichiometry. The balanced chemical equation for the reaction is: 2 Al(s) + 3H₂SO4 (aq) → Al₂(SO4)3 (aq) + 3H₂(g). From the equation, we can see that for every 2 moles of aluminum, 3 moles of sulfuric acid react to produce 3 moles of hydrogen gas. Using the molar masses of aluminum (26.98 g/mol) and sulfuric acid (98.09 g/mol), we can calculate the minimum amount of aluminum required as follows:
25.6 g product × (2 mol Al / 3 mol H₂) × (26.98 g Al / 1 mol Al) = 45.63 g Al
Therefore, the minimum amount of 7.0 m necessary to produce 25.6 g of product is approximately 45.63 g of aluminum.