a) What mass of Na2S2O3 is needed to react with 0.12 g of Cl2? Step 2: Mass to moles Step 3: Mole to Mole (Ratio) Step 4: Mole to Mass

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Md Ariful Islam · Answer 1 · Apr 23, 2024 13:44:01

Solution:

The balanced chemical equation:

Na₂S₂O₃(aq) + 4Cl₂(g) + 5H₂O(aq) --> 2H₂SO₄(aq) + 6HCl(aq) + 2NaCl(aq)

According to the equation: n(Na₂S₂O₃) = n(Cl₂)/4

Step 1 "Mass of Cl₂to moles of Cl₂":

Moles of Cl₂ = n(Cl₂) = Mass of Cl₂ / Molar mass of Cl₂

The molar mass of Cl₂ is 70.9 g mol^-1.

n(Cl₂) = 0.12 g / 70.9 g mol^-1 = 0.00169 mol = 0.0017 mol

Step 2 "Moles of Cl₂to moles of Na₂S₂O₃":

n(Na₂S₂O₃) = n(Cl₂)/4

n(Na₂S₂O₃) = 0.0017 mol / 4 = 0.000425 mol

Step 3 "Moles of Na₂S₂O₃ to mass of Na₂S₂O₃":

Moles of Na₂S₂O₃ = n(Na₂S₂O₃) = Mass of Na₂S₂O₃ / Molar mass of Na₂S₂O₃

The molar mass of Na₂S₂O₃ is 158.11 g mol^-1.

Hence,

Mass of Na₂S₂O₃ = n(Na₂S₂O₃) × M(Na₂S₂O₃)

Mass of Na₂S₂O₃ = 0.000425 mol × 158.11 g mol^-1 = 0.067 g

Mass of Na₂S₂O₃ is 0.067 grams.

Answer: 0.067 g of Na₂S₂O₃ is needed to react with 0.12 g of Cl₂.

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