question 1
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
Question 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
