Respuesta :
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:
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