Question

reaction of sodium thiosulphate pentahydrate with iodine to form sodium tetrathionate

sodium thiosulpahte pentahydrate -0.50 g and iodine 0.24 g

product yield (sodium tetrathionate )- 0.10g

theoretical and perecent yield ?

Answer 1

Reaction between Sodium thipsulphate pentahydrate with iodine is:

I₂ + 2Na₂S₂O₃·5H₂O   5H₂O + 2NaI + Na₂S₄O₆

Mass of Na₂S₂O₃·5H₂O = 0.50 g

Molar mass of Na₂S₂O₃·5H₂O = 248.18 g/mol

Moles of Na₂S₂O₃·5H₂O = mass/ molar mass = 0.50g/(248.18 g/mol) = 0.00201 mol

Mass of I₂ = 0.24 g

Molar mass  of I₂ = 253.8089 g/mol

Moles of I₂ = 0.24 g/(253.8089 g/mol) = 0.000946 mol

From reaction 2 mol of Na₂S₂O₃·5H₂O reacts with 1 mol of I₂

Thus, 0.00201 mol of Na₂S₂O₃·5H₂O reacts with 0.001 mol of I₂

But we have I₂ =  0.000946 mol = limiting reagent

From reaction, 1 mol of I₂ produces 1mol of Na₂S₄O₆

Thus, 0.000946 mol of I₂ produces 1mol of Na₂S₄O₆

Molar mass of Na₂S₄O₆ = 306.2665 g/mol

Mass of  Na₂S₄O₆ = moles * Molar mass =  0.000946 mol * 306.2665 g/mol = 0.290 g

Hence, Theoretical yield = 0.29 g

But actual yield = 0.10 g

Hence, Percent yield = (actual yield/ Theoretical yield )*100% = (0.10/0.29)*100% = 34.53 %