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

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)