Answer to Question #107807 in 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 given can be written as following:

X + H₂SO₄ --> H₂ + XSO₄

The number of moles of H2SO4:

n(H₂SO₄) = C(H₂SO₄) × V(H₂SO₄) = 0.500 mol/dm³ × 100 cm³ = 0.500 mol/L × 0.100 L = 0.05 mol

The number of moles of NaOH use to neutralize H₂SO₄ equals:

n(NaOH) = C(NaOH) × V(NaOH) = 0.500 mol/dm³ × 33.40 cm³ = 0.500 mol/L × 33.40 × 10^-3 L = 0.0167 mol

The neutralization reaction is as following:

H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

As a result, 0.0167 mol of NaOH neutralized 0.0167 / 2 = 0.00835 mol of H₂SO₄.

From here, 0.05 mol - 0.00835 mol = 0.04165 mol of H₂SO₄ reacted with a metal X. As a result, 0.04165 mol of metal X was present.

Molecular weight of metal X equals:

Mr(X) = m(X) / n(X) = 1.00 g / 0.04165 mol = 24 g/mol

Metal X is magnesium (Mg)

Answer: Magnesium (Mg), 24 g/mol