Answer :
Answer : The number of moles of [tex]Fe(OH)_2[/tex] is, 4.225 moles
Solution : Given,
Moles of Fe = 6 moles
Moles of [tex]NiO(OH)[/tex] = 8.45 moles
The given balanced chemical reaction is,
[tex]Fe(s)+2NiO(OH)(s)+2H_2O(l)\rightarrow Fe(OH)_2(s)+2Ni(OH)_2(aq)[/tex]
By Stoichiometry, we conclude that
1 mole of Fe reacts with 2 moles of [tex]NiO(OH)[/tex]
So, 6 moles of Fe will react with = [tex]\frac{2}{1}\times 6=12moles[/tex] of [tex]NiO(OH)[/tex]
From this we conclude that [tex]NiO(OH)[/tex] is considered as a limiting reagent and Fe is the excess reagent.
By Stoichiometry of the given reaction :
2 moles of [tex]NiO(OH)[/tex] produces 1 mole of [tex]Fe(OH)_2[/tex]
So, 8.45 moles of [tex]NiO(OH)[/tex] will produce = [tex]\frac{1}{2}\times 8.45=4.225moles[/tex] of [tex]Fe(OH)_2[/tex]
Therefore, the number of moles of [tex]Fe(OH)_2[/tex] is, 4.225 moles
Answer:
There is 4.225 moles Fe(OH)2 produced.
Explanation:
Step 1: Data given
Number of moles Fe = 6.00 moles
Moles of NiO(OH) = 8.45 moles
Molar mass Fe = 55.85 g/mol
Molar mass NiO(OH) = 91.70 g/mol
Step 2: The balanced equation
Fe(s) + 2 NiO(OH)(s) + 2 H2O(l) → Fe(OH)2(s) + 2 Ni(OH)2(aq)
Step 3: Calculate limiting reactant
For 1 mol Fe we need 2 moles NiO(OH) and 2 moles H2O to produce 1 mol Fe(OH)2 and 2 moles Ni(OH)2
NiO(OH) is the limiting reactant. It will completely be consumed (8.45 moles). Fe is in excess. There will react 8.45 / 2 = 4.225 moles Fe.
There will remain 6.00 - 4.225 moles = 1.775 moles Fe
Step 4: Calculate moles of Fe(OH)2
For 1 mol Fe we need 2 moles NiO(OH) and 2 moles H2O to produce 1 mol Fe(OH)2 and 2 moles Ni(OH)2
For 8.45 moles NiO(OH) we'll have 8.45 /2 = 4.225 moles Fe(OH)2 produced.
There is 4.225 moles Fe(OH)2 produced.