Respuesta :
Ans: Moles of Fe(OH)2 produced is 5.35 moles.
Given reaction:
Fe(s) + 2NiO(OH) (s) + 2H2O(l) → Fe(OH)2(s) + 2Ni(OH)2(aq)
Based on the reaction stoichiometry:
1 mole of Fe reacts with 2 moles of NiO(OH) to produce 1 mole of Fe(OH)2
It is given that there are:
5.35 moles of Fe
7.65 moles of NiO(OH)
Here the limiting reagent is Fe
Therefore, number of moles of Fe(OH)2 produced is 5.35 moles.
Answer:
3.825 mol of Fe(OH)2 are produced .
Explanation:
Reaction:
Fe(s) + 2 NiO(OH)(s) + 2 H2O(l) → Fe(OH)2(s) + 2 Ni(OH)2(aq)
5.35 mol Fe and 7.65 mol NiO(OH) react. For every 1 mol of Fe that reacts 2 mol of NiO(OH) are needed, then for 5.35 mol Fe, 5.35*2 = 10.7 mol of NiO(OH) are needed. Therefore, NiO(OH) is the limiting reactant.
1 mol of Fe(OH)2 is produced when 2 mol of NiO(OH) reacts, then if 7.65 mol NiO(OH) reacts, 7.65/2 = 3.825 mol of Fe(OH)2 are produced .
