Determine the mass of the salt that is formed by the interaction of 520 g of acetic acid with potassium.

Given:
m (CH3COOH) = 520 g
To find:
m (CH3COOK) =?
Decision:
1) We draw up the equation of the chemical reaction, let the mass of the salt be x g.
2K + 2CH3COOH = 2CH3COOK + H2
2) Find the molecular weights of acetic acid and salt.
M (CH3COOH) = 12 + (1 x 3) + 12 + 16 + 16 + 1 = 60 g / mol
M (CH3COOK) = 12 + (1 x 3) + 12 +16 + 16 + 39 = 98 g / mol
3) 520 g: 60 g / mol = x g: 98 g / mol
x = (520 x 98): 60 = 849.33
Answer: m (CH3COOK) = 849.33 g.