Answer to Question #107805 in Organic Chemistry for Lala

Question

a 1.00 g sample of a metal x (that is known to form x+2 ions) was added to a 0.100cm3 of 0.500mol dm3 h2so4. after all the metal head reacted, the remaining acid required 33.40 cm3l of 0.500mol dm3 naoh solution for neutralization.calculate the molar mass of the metal and identify the element.

Answer 1

The reaction between X and H₂SO₄:

X + H₂SO₄ --> H₂ + XSO₄

To identify the molecular weight of metal X, the number of moles should be calculated. As metal X interacted completely with the acid, the number of moles of H₂SO₄ that reacted with the metal X should be found.

Total number of moles of the acid equals:

n(H₂SO₄) = C(H₂SO₄) × V(H₂SO₄) = 0.500 mol/dm³ × 100 cm³ = 0.05 mol

The remaining acid was neutralized with NaOH in the following reaction:

H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

As a result, the number of moles of NaOH used to neutralize H₂SO₄ equals:

n(NaOH) = C(NaOH) × V(NaOH) = 0.500 mol/dm3 × 33.40 cm3 = 0.017 mol

As the ratio between the acid and base equals 1/2, 0.0167 mol of NaOH neutralized 0.017 / 2 = 0.0085 mol of H₂SO₄.

From here, 0.05 mol - 0.0085 mol = 0.0415 mol of H₂SO₄ reacted with a metal X. As a result, 0.0415 mol of metal X reacted with the acid.

Molecular weight of metal X equals:

Mr(X) = m(X) / n(X) = 1.00 g / 0.0415 mol = 24.1 g/mol

Metal X is magnesium (Mg)

Answer: 24.1 g/mol, Mg.