Question

5. Based on the following reaction: _Fea(SO4)3 +_ KOH > _K2SO4 + _Fe(OH)

Answer 1

Answer:

a. Fe₂(SO₄)₃  + 6 KOH  →  3 K₂SO₄  +   2Fe(OH)₃

b. 2 mol of KOH

c. 8 mol of Iron(II) hydroxide

Explanation:

This is the balanced reaction

Fe₂(SO₄)₃  + 6 KOH  →  3 K₂SO₄  +   2Fe(OH)₃

b. Ratio is 3:6 so,

For 3 mol of potassium sulfate, I needed 6 mol of KOH

For 1 mol of potassium sulfate I would need (1 .6 )/3 = 2 mol of KOH

c. 1 mol of iron(III) sulfate, produce 2 mol of Iron(II) hydroxide

4 mol of iron(III) sulfate, will produce (4x2), 8 mol of Iron(II) hydroxide

Answer 2

Answer:

a) Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3

b) 2 moles of KOH

c) 8 moles of Fe(OH)3

Explanation:

Step 1: The balanced equation

Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3

Step 2: Data given

Molar mass of K2SO4 = 174.26 g/mol

Molar mass of KOH = 56.11 g/mol

Molar mass of Fe(OH)3 = 106.87 g/mol

Molar mass of Fe2(SO4)3 = 399.88 g/mol

How many mole of potassium hydroxide (KOH) would be needed to create 1 mole of K2SO4 ?

Step 3: Calculate moles of potassium hydroxide (KOH)

For 1 mol Fe2(SO4)3 we need 6 moles of KOH to produce 3 moles of K2SO4 and 2 moles of Fe(OH)3

To produce 1 mol of K2SO4 we need 2*1 = 2 moles of KOH

How many mole of Fe(OH)3, would be produced from reacting 4 mole of Fe2(S04)3 ?

Step 4: Calculate moles of Fe(OH)3

For 1 mol Fe2(SO4)3 we need 6 moles of KOH to produce 3 moles of K2SO4 and 2 moles of Fe(OH)3

When 4 moles of Fe2(SO4)3 will be consumed, there will be produced 2*4 = 8 moles of Fe(OH)3