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# Answer to Question #166699 in Inorganic Chemistry for Richard

Question #166699

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?

1
2021-02-26T05:55:41-0500

"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)

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