Potassium iodide reacts with lead(ii) nitrate in the following precipitation reaction:
2ki(aq)+pb(no3)2(aq)→2kno3(aq)+pbi2 (s)

what minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 155.0 ml of a 0.122 m lead(ii) nitrate solution?

0.1891 L minimum volume of 0.200 M potassium iodide solution is required.

Explanation:

Considering:

Or,

Given :

For lead(II) nitrate solution :

Molarity = 0.122 M

Volume = 155.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 155.0×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles of lead(II) nitrate  = 0.01891 moles

According to the given reaction:

1 mole of lead(II) nitrate reacts with 2 moles of potassium iodide

0.01891 mole of lead(II) nitrate reacts with 2*0.01891 moles of potassium iodide

Moles of potassium iodide = 0.03782 moles

Also,

Molarity = 0.200 M

Volume = ?

0.1891 L minimum volume of 0.200 M potassium iodide solution is required.

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: