Answer to Question #343939 in General Chemistry for Theogene

Question

An experiment was carried out to determine the mass of Sodium oxide in a solution sample. The sample was dissolved in water to make 200 cm³ of Sodium hydroxide. This solution required 500 cm³of Sulphuric acid to just neutralize it. The concentration of Sulphuric acid was 0.1mol/dm³ (0.1M)

Answer 1

Na₂O + H₂O => 2NaOH

2NaOH + H₂SO₄ => Na₂SO₄ + 2H₂O

1. We find amount substance(n, mol) of Sulphuric acid:

n (H₂SO₄) = V * C_M

n – amount substance of sulphuric acid(mol);

V – solution volume of sulphuric acid = 500 cm3 = 50 dm³ = 50 litres;

C_M – concentration of sulphuric acid = 0,1 mol/l .

n = 50 * 0,1 = 5 mol.

2. We find amount substance of sodium hydroxide. According to the above reaction, 1mol H₂SO₄ reacts with 2mol NaOH. How many moles NaOH react with 5mol H₂SO₄ :

1 mol H₂SO₄ — 2 mol NaOH

5 mol H₂SO₄ — X mol NaOH

X= 5*2/1 = 10 mol NaOH

3.We find the mass of Sodium oxide (Na₂O)

According to the first reaction, 1 mol (62 g) of Na₂O produces 2 mol of NaOH:

62 gr Na₂O — 2 mol NaOH

X gr Na₂O — 10 mol NaOH

X = 62 * 10 / 2 = 310 gr

ANSWER: 310 gr sodium oxide (Na₂O)