Given 2 grams feso4 and excess naoh, calculate how much fe(oh)2 can be produced? given 2 grams fe(no3)3 and excess naoh, calculate how much fe(oh)3 can be produced?

The much  of  Fe(OH)2 that can be produced is  1.188  grams

calculation

step 1:  write the equation  for reaction

FeSO₄ +2NaOH   Fe(OH)₂  +Na₂SO₄

Step 2:  calculate the  moles of  FeSO₄

moles = mass/molar mass

from periodic table the  molar mass of FeSo₄  is 152 g/mol

= 2 g/152 g/mol=0.0132  moles

step 3:use  the mole  ratio  of  to determine the  moles of Fe(OH)₂

from equation above FeSO4: Fe(OH)₂  is  1:1  therefore the  moles of Fe(OH)₂  is also  = 0.0132  moles

step 4:  find mass  of Fe(OH)₂

mass = moles x  molar mass

From periodic table the molar mass of Fe(OH)2  = 90 g/mol

mass =0.0132  moles  x 90  g/mol  = 1.188 grams  of Fe(OH)₂

The  much  of Fe(OH)3  that can be produced  is 0.888 grams

calculation

Step 1: write  the equation for reaction

Fe(NO₃)₃ +3NaOH →   Fe(OH)₃  +3NaNO₃

Step 2:  find the  moles of fe(NO₃)₃

moles = mass/molar  mass

from  periodic table the molar mass of Fe(NO₃)₃  is = 242  g/mol

moles = 2 g/242 g/mol  =0.0083  moles

Step 3: use  the  mole ratio to determine the  moles of Fe(OH)₃

 from equation above Fe(NO₃)₃: Fe(OH)₃  is 1:1  therefore  the  moles  of Fe(OH)₃ is also =0.0083 moles

step 4: find mass  of Fe(OH)₃

mass=  moles x  molar mass

from periodic table the  molar mass of Fe(OH)₃ =107 g/mol

mass = 0.0083  moles x 107 g/mol =0.888 grams