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)

true

explanation:

who knows

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :