Question

Al2(so3)3 6 naoh ------> 3 na2so3 2 al(oh)3

if 10.0 g of al2(so3)3 is reacted with 10.0 g of naoh, determine the limiting reagent
determine the number of grams of na2so3 produced.
determine the number of grams of excess reagent left over in the reaction

Answer 1

To solve the question, we calculate the moles of Al2(SO3)3 and NaOH, identify the limiting reagent using stoichiometry of the balanced equation, determine the mass of Na2SO3 produced, and calculate the mass of excess reagent remaining.

Step-by-step explanation:

The question involves a chemical reaction where aluminum sulfite (Al2(SO3)3) reacts with sodium hydroxide (NaOH) to produce sodium sulfite (Na2SO3) and aluminum hydroxide (Al(OH)3).

To determine the limiting reagent and the amount of product formed, we need to:

1. Calculate the number of moles of each reactant using their molar masses.
2. Determine which reactant produces the least amount of product by using the stoichiometry of the balanced chemical equation.
3. Convert the number of moles of product (determined by the limiting reagent) to grams using its molar mass.
4. Calculate the amount of excess reagent left over by subtracting the moles of excess reactant used from the initial moles present.

By performing these calculations, we can identify the limiting reagent and the mass of Na2SO3 produced, as well as the mass of the excess reagent remaining after the reaction.