Then if I have 210.1 g barium chloride with 45g potassium...how many grams of potassium chloride will I produce? Do the stoich

I need help please is for today ):

  85.8 g

Explanation:

The molar mass of BaCl₂ is 208.233 g/mol, so 210.1 g is about ...

  210.1/208.233 ≈ 1.009 mol

The molar mass of K is about 39.089 g/mol, so 45 g is about ...

  45/39.089 ≈ 1.1509 mol

The balanced equation seems to be ...

  BaCl₂ + 2K  ↔ 2KCl + Ba

This reaction requires 2 moles of potassium for each mole of barium chloride, so the available potassium limits the reaction. The result is as many moles of potassium chloride as there are moles of potassium.

The 1.1509 moles of KCl, at 74.551 g/mol will have a mass of about 85.8 g.

The reaction produces 85.8 g of KCl.

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:

fe 55.85 g/mol and h2o 18.02 g/mol.