Find the mass of iron hydroxide (lll) obtained by oxidation of 8 mol of iron hydroxide (ll),

Find the mass of iron hydroxide (lll) obtained by oxidation of 8 mol of iron hydroxide (ll), if the yield of iron hydroxide (lll) is 90%

To solve the problem, we compose the equation of the process:

4Fe (OH) 2 + O2 + 2H2O = 4Fe (OH) 3 – OBP, iron hydroxide (3) was obtained;
Calculations:
M Fe (OH) 3 = 106.8 g / mol.

3. Proportion:

8 mol Fe (OH) 2 – X mol Fe (OH) 3;

-4 mol -4 mol from here, X mol Fe (OH) 3 = 8 * 4/4 = 8 mol.

Find the mass of the product:
m Fe (OH) 3 = Y * M = 8 * 106.8 = 854.4 g (theoretical weight);

m Fe (OH) 3 = 0.90 * 854.4 = 768.96 g (practical weight).

Answer: Received iron (3) hydroxide weighing 768.96 g