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 Cl2 to 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 Cl2 to 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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