Answer :
The answer is: 4.00 mol of Fe(OH)2 is produced.
Balanced chemical reaction:
Fe(s) + 2NiO(OH)(s) + 2H₂O(l) → Fe(OH)₂(s) + 2Ni(OH)₂(aq).
n(Fe) = 5.00 mol; amount of iron.
n(NiO(OH)) = 8.00 mol; amount of nickel oxide hydroxide, limiting reactant.
From chemical reaction: n(NiO(OH)) : n(Fe(OH)₂) = 2 : 1.
n(Fe(OH)₂) = 8 mol ÷ 2.
n(Fe(OH)₂) = 4 mol; amount of iron(II) hydroxide.
Answer:
4.0 moles of Fe(OH)₂.
Explanation:
- It is a stichiometry problem.
- The balanced equation: Fe + 2NiO(OH) + 2H₂O → Fe(OH)₂ + 2Ni(OH)₂,
- It is clear that 1.0 mole of Fe reacts with 2.0 moles of NiO(OH) and 2.0 moles of H₂O to produce 1.0 mole of Fe(OH)₂ and 2.0 moles of Ni(OH)₂.
- The given data: 5.00 moles of Fe reacts with 8.00 mol of NiO(OH).
- From the stichiometry; 4.0 moles of Fe (1.0 mole will be in excess) reacts completely with 8.0 moles of NiO(OH) and produce 4.0 moles of Fe(OH)₂.