A 4.476 g sample of a petroleumproduct was burnt in a tubefurnace, and the SO 2
ID: 677785 • Letter: A
Question
A 4.476 g sample of a petroleumproduct was burnt in a tubefurnace, and the SO2 produced was collected in 3%H2O2. Reaction: SO2 (g) +H2O2--> H2SO4. A 25.00 mL portion of 0.00923 M NaOH was introduced into thesolution of the H2SO4, following which theexcess base was back titrated with 13.33 mL of 0.01007 M HCL.Calculate the part per million sulfur in the sample. A 4.476 g sample of a petroleumproduct was burnt in a tubefurnace, and the SO2 produced was collected in 3%H2O2. Reaction: SO2 (g) +H2O2--> H2SO4. A 25.00 mL portion of 0.00923 M NaOH was introduced into thesolution of the H2SO4, following which theexcess base was back titrated with 13.33 mL of 0.01007 M HCL.Calculate the part per million sulfur in the sample.Explanation / Answer
Moles of NaOH = Molarity * volume=0.00923 M * 0.025 L =0.00023 moles Excess NaOH titrated with 13.33 m L of 0.01007 M HCl. Moles of HCl = 0.01007 M * 0.01333 L =0.000134 moles So moles of NaOH consumed by H2SO4 = 0.00023 - 0.000134 =0.000096 moles Since H2SO4 is a diprotic acid, moles of acid = 0.000096 / 2 =0.000048 moles Given SO2 (g) + H2O2-->H2SO4. Since 1 mole of SO2 gives 1 mole of sulfuric acid, Moles of SO2 = 0.000048 moles. Sulfur on combustion gives Sulfur dioxide S+ O2 ------> SO2 So moles of sulfur = 0.000048 So mass of sulfur = Moles * molar mass =0.000048 * 32.07 g/mol = 0.00153 g Total mass of sample = 4.476 g So concentration of sulfur in ppm = ( Mass of sulfur / Mass ofsample ) * 10 6 =( 0.00153 g/ 4.476 g) * 10 6 =341.82 ppm =341.82 ppm
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