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?

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