Question

1. Given the following unbalanced equation:

Al(SO3)3 + NaOH
---> Na2SO3 + Al(OH)3

If you start with 389.4 grams of Al(SO3)3 and 200 grams of NaOH ?

Identify the limiting reactant:

Identify the reactant in excess and by how much:

How many grams of Na2SO, will be produced:

If you perform the same reaction but get 212.4 grams of Na2SO3, what is
your percent yield?

Answer 1

Limting reactant: NaOH

Excess reactant: Al₂(SO₃)₃ by 104.9 grams.

315 g of Na₂SO₃ are produced in the reaction

67.4 % is the percent yield.

Step-by-step explanation:

Balaced equation is:

Al₂(SO₃)₃ + 6NaOH → 3Na₂SO₃ + 2Al(OH)₃

We convert the mass of each reactant to moles, to identify the limiting reactant:

389.4 g . 1mol/ 342.15g = 1.14 moles of sulfite

200 g . 1mol / 40g = 5 moles

1 mol of aluminum sulfite react to 6 moles of NaOH, according to stoichiometry.

If we have 1.14 moles of sulfite, we need (1.14 . 6)/1 = 6.82 moles

We only have 5 mol of NaOH, so this is the limiting reactant.

Then the aluminum sulfite is the excess.

6 moles of NaOH react to 1 mol of sulfite

Then, 5 moles of NaOH may react to (5 . 1)/6 = 0.83 moles.

1.14 mol - 0.83 mol = 0.31 moles remains after the reaction goes complete.

We convert to mass: 0.31 mol . 342.15g/mol = 104.9 g

Then, we work with the limiting reactant. Stoichiometry is 6:3 (2)

6 moles of NaOH can produce 3 moles of sodium sulfite

Then 5 moles will produce (5 . 3)/6 = 2.5 moles

We convert moles to mass: 2.5 mol . 126 g/mol = 315 g.

Percent yield = (Yield produced/Thoeretical yield) . 100

Percent yield = (212.4g /315g) . 100 = 67.4 %