Question

(Balanced) Equation- 2NaOH + FeCl2 (react to produce) 2NaCl + Fe(OH)2 Question- If 12.75g of NaOH and 20.20g of FeCl2 reacted to produce 18.63g of NaCl, how much Fe(OH)2 was produced?

Answer 1

`0.159molFe(OH)_2`
`14.31gFe(OH)_2`

Explanation:

First, convert the grams into moles:
`12.75 g NaOH * (1 mol NaOH)/(40 g NaOH) = 12.75/40 ~~ 0.319molNaOH`
`20.20 g FeCl_2 * (1 mol FeCl_2)/(1270 g FeCl_2) = 20.2/127 ~~ 0.159molFeCl_2`
`18.63 g NaCl * (1 mol NaCl)/(58.5gNaCl) = 18.63/58.5 ~~ 0.318molNaCl`

Then, determine which is the limiting reactant by using the balanced equation(KhanAcademy link here ):
`0.319molNaOH * (1molFe(OH)_2)/(2molNaOH) ~~ 0.159molFe(OH)_2`
`0.159molFeCl_2 * (1molFe(OH)_2)/(1molFeCl_2 ) ~~ 0.159molFe(OH)_2`
`0.318molNaCl * (1molFe(OH)_2)/(2molNaCl) ~~ 0.159molFe(OH)_2`

As they would theoretically produce the same amount, they are all limiting reactants so any can be used for the next step.

Finally, take the amount of moles `Fe(OH)_2` that could be produced and convert into grams:
`0.159molFe(OH)_2 * (90gFe(OH)_2)/(1molFe(OH)_2) = 14.31gFe(OH)_2`