Alaboratory 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

 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)

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