To 40 g of an 8% solution of sodium hydroxide alkali, a solution containing 0.2 mol

To 40 g of an 8% solution of sodium hydroxide alkali, a solution containing 0.2 mol of ferrous sulfate (2) was added, calculate the mass of the sediment.

1. Let’s write the equation of interaction of sodium hydroxide with iron (II) sulfate:

2NaOH + FeSO4 = Fe (OH) 2 ↓ + Na2SO4;

2.Calculate the mass of sodium hydroxide:

m (NaOH) = w (NaOH) * m (NaOH solution) = 0.08 * 40 = 3.2 g;

3. find the chemical amount of sodium hydroxide:

n (NaOH) = m (NaOH): M (NaOH);

M (NaOH) = 23 + 16 + 1 = 40 g / mol;

n (NaOH) = 3.2: 40 = 0.08 mol;

4.the deficiency will be sodium hydroxide, therefore, according to its data, we will find the amount of iron (II) hydroxide:

n (Fe (OH) 2) = n (NaOH): 2 = 0.08: 2 = 0.04 mol;

5.Calculate the mass of the sediment:

m (Fe (OH) 2) = n (Fe (OH) 2) * M (Fe (OH) 2);

M (Fe (OH) 2) = 56 + 2 * 17 = 90 g / mol;

m (Fe (OH) 2) = 0.04 * 90 = 3.6 g.

Answer: 3.6 g.



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