What volume of a 0.88 M solution can be made using 130. Grams of FeCI^2?

1.16L can be made

Explanation:

Molarity = Mol / Volume

Volume = Mol / Molarity

Let's determine the moles of salt, with that mass:

130 g FeCl₂ . 1mol / 126.75 g = 1.02 moles of FeCl₂

Volume = 1.02 mol / 0.88 mol/L → 1.16L

Three (3) => 2Na⁺ and 1SO₄²⁻

Explanation:

Na₂SO₄ => 2Na⁺ + SO₄²⁻

Note: one can easily determine the cation-anion separation during aqueous ionization if one reads the formula left to right => place an imaginary line after the 1st metal of the molecular formula => left of line is cation and right of line anion.

Na₂SO₄ =>  Na₂ | SO₄ => 2Na⁺ + SO₄²⁻  (3 ions)

CaCO₃ => Ca | CO₃ => Ca₂⁺ + CO₃²⁻  (2 ions)

Ni(NO₃)₂ => Ni | (NO₃)₂ => Ni²⁺ + 2NO₃⁻ (3 ions)

Ca₃(PO₄)₂ => Ca₃ |  (PO₄)₂ => 3Ca²⁺ + 2PO₄³⁻  (5 ions)