Final answer:
The theoretical yield of Al2(SO4)3 in grams can be calculated using stoichiometry. Given the mass of aluminum, we can convert it to moles and then use the balanced equation to determine the moles of Al2(SO4)3. Finally, the moles can be converted to grams using the molar mass of Al2(SO4)3.
Step-by-step explanation:
The theoretical yield in grams of Al2(SO4)3 for this reaction can be calculated using stoichiometry. The balanced equation for the reaction is 2 Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g). From the equation, we can see that 2 moles of Al react with 3 moles of H2SO4 to produce 1 mole of Al2(SO4)3. To calculate the theoretical yield, we need to know the amount of Al that reacts.
Given that a 2.00 g piece of aluminum completely reacts, we can convert the mass of Al to moles using its molar mass. The molar mass of Al is 26.98 g/mol, so:
moles of Al = 2.00 g / 26.98 g/mol = 0.0741 mol
Therefore, the theoretical yield of Al2(SO4)3 is:
(0.0741 mol Al) x (1 mol Al2(SO4)3 / 2 mol Al) x (342.14 g/mol Al2(SO4)3) = 12.6 g Al2(SO4)3