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?

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.

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.