Question

A laboratory technician combined sodium hydroxide with excess iron(II) nitrate. A reaction took place according to this chemical equation: 2NaOH + Fe(NO3)2 → NaNO3 + Fe(OH)2.
The reaction produced 3.70 grams of iron(II) hydroxide.

Assuming the reaction came to completion, what was the initial mass of sodium hydroxide? Use the periodic table.

A. 1.6 g
B. 2.0 g
C. 3.3 g
D. 4.0 g

Answer 1

 The initial  mass  of sodium hydroxide  is  3.3 g (answer C)

 calculation

Step 1 : find the  moles of iron (ii) hydroxide (  Fe(OH)₂

moles =  mass÷  molar mass

from periodic table the  molar mass of Fe(OH)₂  = 56 + [16 +1]2  = 90 g/mol

moles  is therefore = 3.70 g÷ 90 g/mol = 0.041 moles

Step 2:  use the mole ratio to  calculate the moles of  sodium hydroxide (NaOH)

   from given equation  NaOH : Fe(OH)₂    is 2 :1

therefore the moles of NaOH = 0.041 x 2 = 0.082 moles

Step 3: find  mass of NaOH

mass = moles x molar mass

from the periodic table the  molar mass of NaOH = 23 +16 +1  = 40 g/mol

mass  = 0.082  moles x 40 g/mol = 3.3 g ( answer C)