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 6.00 mol Fe and 8.45 mol NiO(OH) react?

Answer 1

There is 4.225 moles Fe(OH)2 produced.

Step-by-step 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.

Answer 2

Answer : The number of moles of
`Fe(OH)_2` is, 4.225 moles

Solution : Given,

Moles of Fe = 6 moles

Moles of
`NiO(OH)` = 8.45 moles

The given balanced chemical reaction is,

`Fe(s)+2NiO(OH)(s)+2H_2O(l)\rightarrow Fe(OH)_2(s)+2Ni(OH)_2(aq)`

By Stoichiometry, we conclude that

1 mole of Fe reacts with 2 moles of
`NiO(OH)`

So, 6 moles of Fe will react with =
`(2)/(1)* 6=12moles` of
`NiO(OH)`

From this we conclude that
`NiO(OH)` is considered as a limiting reagent and Fe is the excess reagent.

By Stoichiometry of the given reaction :

2 moles of
`NiO(OH)` produces 1 mole of
`Fe(OH)_2`

So, 8.45 moles of
`NiO(OH)` will produce =
`(1)/(2)* 8.45=4.225moles` of
`Fe(OH)_2`

Therefore, the number of moles of
`Fe(OH)_2` is, 4.225 moles