What mass of salt will be obtained by reacting with zinc oxide 2 liters of ethanic acid, if the product yield is 80%.

Let’s implement the solution:

Let’s compose the process equation according to the problem statement:
V = 2 l X g -? W = 80%

2СН3СООН + ZnO = (CH3COO) 2Zn + H2O – exchange, zinc acetate was obtained.

2.Calculation:

M (CH3COOH) = 60 g / mol;

M (zinc acetate) = 183.3 g / mol.

Proportions:
1 mol of gas at normal level – 22.4 liters;

X mol (CH3COOH) – 2 liters from here, X mol (CH3COOH) = 1 * 2 / 22.4 = 0.09 mol;

0.09 mol (CH3COOH) – X mol (zinc acetate);

-2 mol -1 mol hence, X mol (zinc acetate) = 0.09 * 1/2 = 0.045 mol.

Find the mass of salt:
m (zinc acetate) = Y * M = 0.045 * 183.3 = 8.25 g (theoretical weight);

m (zinc acetate) = 0.80 * 8.25 = 6.6 g (practical weight).

Answer: The mass of zinc acetate is 6.6 g