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

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  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.

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explanation: