If 22.5 g of NaOH is added to 0.800 L of 1.00 M Zn(NO₃)₂, what mass in grams of Zn(OH)₂ will be formed in the following precipitation reaction? 2 NaOH(aq) + Zn(NO₃)₂(aq) → Zn(OH)₂ (s) + 2 NaNO₃ (aq)

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:

dynamic equilibrium is the answer.

what is the question here. i think you messed up.