For the reaction 4fecl2(aq) + 3o2(g) → 2fe2o3(s) + 4cl2(g), what volume of a 0.945 m solution of fecl2 is required to react completely with 4.32 ×1021 molecules of o2?

0.01M

Explanation:

Given paramters:

Volume of FeCl₂ = ?

Concentration = 0.945M

Number of molecules of O₂ = 4.32 x 10²¹molecules

Solution:

The balanced reaction equation is given below:

           4FeCl₂ + 3O₂ → 2Fe₂O₃ + 4Cl₂

Now, to solve the problem we use mole relationship between the compounds.

We work from the known to the unknown compounds. Here, the known is the oxygen gas because from the given parameters we can estimate the number of moles of the reacting gas.

  number of moles of O₂ =

number of moles of O₂ =  

number of moles of O₂ = 0.007mole

now, from the reaction equation we find the number of moles of the FeCl₂:

   3 mole of O₂ reacted with 4 moles of FeCl₂

0.007mole of O₂ reacted with = 0.0096mole O₂

Now to find the volume of FeCl₂ we use the expression below;

       volume of FeCl₂ =

                                   =  

                                    = 0.01L

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

c. electrical charges of ionic bonds.