2Fe(SO4)3+6koh--->2Fe(OH)3+3k2(SO4)

if you started with 12.3 g of potassium hydroxide and a sufficient amount of iron III sulfate how many grams of iron III hydroxide will form?

mols KOH = grams/molar mass

mols KOH = ?mols Fe(OH)3---Use the coefficients in the balanced equation.
g Fe(OH)3 = mols x molar mass.

To find the number of grams of iron III hydroxide that will form, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.

First, we need to convert the mass of potassium hydroxide (KOH) to moles. The molar mass of KOH is 56.11 g/mol.

12.3 g KOH * (1 mol KOH / 56.11 g KOH) = 0.2194 mol KOH

According to the balanced chemical equation, the ratio between KOH and Fe(OH)3 is 6:2.

0.2194 mol KOH * (2 mol Fe(OH)3 / 6 mol KOH) = 0.0731 mol Fe(OH)3

Now, we can convert moles of Fe(OH)3 to grams using its molar mass. The molar mass of Fe(OH)3 is 106.87 g/mol.

0.0731 mol Fe(OH)3 * (106.87 g Fe(OH)3 / 1 mol Fe(OH)3) = 7.81 g Fe(OH)3

Therefore, 7.81 grams of iron III hydroxide will form in this reaction.