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?

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.