Final answer:
To find the limiting reagent in the reaction Al2(SO3)3 + 6 NaOH → 3 Na2SO3 + 2 Al(OH)3, convert the given masses of reactants into moles and compare these with the stoichiometric ratios from the balanced equation. The reagent that yields fewer moles of product is the limiting reagent, which is NaOH in this case.
Step-by-step explanation:
The question asks to determine the limiting reagent when 10.0 g of Al2(SO3)3 is reacted with 10.0 g of NaOH in the reaction: Al2(SO3)3 + 6 NaOH → 3 Na2SO3 + 2 Al(OH)3. To solve this, we must first convert the masses to moles. For Al2(SO3)3:
Molar mass = (2×26.98) + (3×32.06) + (3×24×16.00) g/mol
Moles of Al2(SO3)3 = 10.0 g / molar mass of Al2(SO3)3. For NaOH:
Molar mass = 22.99 + 15.99 + 1 g/mol
Moles of NaOH = 10.0 g / molar mass of NaOH.
Next, determine the mole ratio required by the balanced equation and compare it to the mole ratio of the reactants. The reactant that produces fewer moles of product is the limiting reagent. For this reaction, NaOH is determined to be the limiting reagent based on stoichiometric calculations.