Answer:
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.
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 %
