# Question b9efe

Apr 10, 2014

I assume that the concentration of the Na₂S₂O₃ was 0.1 mol•dm⁻³.

The equations are:

BaCrO₄ → Ba²⁺ + CrO₄²⁻

2CrO₄²⁻ + 6I⁻ + 16H⁺ → 2Cr³⁺ + 3I₂ + 8H₂O

I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

Moles of I₂ = 0.030 dm³ S₂O₃²⁻ ×

(0.1" mol S₂O₃²⁻")/(1" dm³ S₂O₃²⁻") × (1" mol I₂")/(2" mol S₂O₃²⁻"# = 0.002 mol I₂

Moles of CrO₄²⁻ =0.002 mol I₂ × $\left(2 \text{ mol CrO₄²⁻")/(3" mol I₂}\right)$ = 0.001 mol CrO₄²⁻

[CrO₄²⁻] = $\left(0.001 \text{ mol")/(0.030" dm³}\right)$ = 0.03 mol/dm³

Mass of BaCrO₄ = 0.001 mol CrO₄²⁻ ×

$\left(1 \text{mol BaCrO₄")/(1" mol CrO₄²⁻") × (253.3" g BaCrO₄")/(1" mol BaCrO₄}\right)$ = 0.3 g BaCrO₄

Mass of BaSO₄ = 0.929 g – 0.3 g = 0.7 g

Moles of SO₄²⁻ = 0.7 g BaSO₄ ×

$\left(1 \text{ mol BaSO₄")/(233.4" g BaSO₄") × (1" mol SO₄²⁻")/(1" mol BaSO₄}\right)$ = 0.003 mol SO₄²⁻

[SO₄²⁻] = $\left(0.003 \text{ mol")/(0.030" dm³}\right)$ = 0.0010 mol/dm³

Note: I have calculated the answer to only 1 significant figure. That is all you gave me for the concentration of the Na₂S₂O₃. If you need more precision, you will have to recalculate.