Question

In 1901, Thomas Edison invented the nickel-iron battery. The following reaction takes place in the battery. Fe(s) + 2 NiO(OH)(s) + 2 H2O(l) Fe(OH)2(s) + 2 Ni(OH)2(aq) How many mole of Fe(OH)2, is produced when 5.35 mol Fe and 7.65 mol NiO(OH) react?

Answer 1

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 .

Answer 2

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.