Answer:
0.191M of Na₂C₂O₄ is the concentration of the original Na₂C₂O₄ solution
Explanation:
The reaction of potassium dichromate, K₂Cr₂O₇ with sodium oxalate, Na₂C₂O₄ in the presence of acid H⁺ is:
K₂Cr₂O₇ + 3Na₂C₂O₄ + 14H⁺ → 2Cr³⁺ + 6CO₂ + 7H₂O + 6Na⁺ + 2K⁺
<em>Thus, 1 mole of K₂Cr₂O₇ reacts with 3 moles of Na₂C₂O₄</em>
Moles of 2.34g of K₂Cr₂O₇ (Molar mass: 294.185g/mol):
2.34g K₂Cr₂O₇ ₓ (1mol / 294.185g) = 7.954x10⁻³ moles K₂Cr₂O₇
In 250mL = 0.250L:
7.954x10⁻³ moles K₂Cr₂O₇ / 0.250L = 0.0318M K₂Cr₂O₇
Moles in 35.7mL = 0.0357L of this solution are:
0.0357L ₓ (0.0318mol / L) = <em>1.136x10⁻³ moles K₂Cr₂O</em>₇ in solution. As 1 mole of K₂Cr₂O₇ reacts with 3 moles of Na₂C₂O₄, to titrate the moles of K₂Cr₂O₇ in solution you need:
1.136x10⁻³ moles K₂Cr₂O₇ × (3 moles Na₂C₂O₄ / 1 mole K₂Cr₂O₇) =
<em>3.408x10⁻³ moles of Na₂C₂O₄</em>
In 17.8mL = 0.0178L:
3.408x10⁻³ moles of Na₂C₂O₄ / 0.0178L =
<h3>0.191M of Na₂C₂O₄ is the concentration of the original Na₂C₂O₄ solution</h3>
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