Answer to Question #166699 in Inorganic Chemistry for Richard

Question

Sulfur (1400 kg) and pearl caustic soda (950 kg) are reacted to form sodium polysulphide (Na2Sx) to a total of 4000 L (balanced by H2O). What percentages (%) of total sulphur (S), sodium sulphide (Na2S), sodium thiosulphate (Na2S2O3) and sodium sulfur compounds are in solution?

Answer 1

$4S+6NaOH\rightarrow Na_2S_2O_3+2Na_2S+3H_2O$

Moles of Sulphur = $\dfrac{1400}{32}=43.75 kmol$

Moles of NaOH = $\dfrac{950}{40}=23.75kmol$

From reaction,

6 moles of NaOH react with 4 moles of S

23.75 kmol of NaOH reacts with 15.833 kmol of S

Moles of S left = 43.75 - 15.833 = 27.91 kmol

Total volume of Solution = 4000 L

Moles of Na₂S₂O₃ formed = 3.95 kmol

Moles of Na₂S formed = 7.91 kmol

Moles of H₂O formed = 11.87 kmol

Mass of S left = $27.91\times32=893.12kg$

Mass of Na₂S₂O₃ formed = $3.95\times158=624.1kg$

Mass of Na₂S formed = $7.91\times78=616.98 kg$

Mass of H₂O formed = $213.66kg$

Mass/Volume percentage of components,

$S=\dfrac{893.12}{4000}\times100=22.328\%$

$Na_2S_2O_3=\dfrac{624.1}{4000}\times 100=15.6\%$

$Na_2S=\dfrac{616.98}{4000}\times100=15.42\%$

$H_2O=100-(22.328+15.6+15.42)=46.652\%$ (since H₂O is added externally also)