Question

Chapter 15: EOC Question 55 p (References A4980 g sample of a mixture of sodium carbonate and sodium chloride is dissolved in 25.40 ml. of 0.792 M HCl. After treatment of the sample, some acid remai 1f219 ml. of 0.116 M NaOH were required to titrate the excess hydrochloric acid, how many moles of sodium carbonate were present in the original sample? Question 56 1 pr mol Nasco Question ST 1 pt Question 58 fpt What is the percent composition of the original sample? % NaCO3 Question 50 pe Question 600 pe Naci Question 1 1 Sub Answer Try Another Wersion Question 2 pe Question pr Question 64 pl Question 65 pl Question o f pr Question 671 pt Question 631 pc Questione pe Question 70 pe Dua Maya

Answer 1

solution Moss of Sample (Na₂CO3 + Nall) = 4.980g. Hel = 0.792M, 25. 40ML = 25.40 2 Sime = 1000 no. of moles of HU= Molarityx volume in = 0.792 = 0.792 med mol x 25.40K = 0.020lmole 25.000 e NaOH = 0.116 M, 2109mL = 21092 no. of moles of NaOta dellb mol x 21.9K = 0.0025 wale 1000 K 1000 As Hill & NaOH Wall + H₂O. » Imole of NaOH quench I mole of Hel. 2) 0.0025 mole of daoH will quench 0.00 25 mole of Hll. - no. of moles of Hel quenched by NaOH = 0.002 moles. Remaining no. of moles of Hill = 0.0201-0.0025 = 0.0176 moles 0.0176 moles of Hell quenched by Na2CO3. Na₂CO3 +2HU - aNall + CO2+ H₂O » 2 mole of Hel quenched by 1 mole of a CO₂ = 0.0176 moles of Hill quenched by ± x 0.0176 mole of Na, coz = 0.0088 mole of NaCO3 » No. of moles of Naz CO₂ = 0.0088 mole. = 0.009 mole Holar mass of Na₂CO₃ = 106glomet - Mass of Naz (O3 = no. of moles x Molor mass NaCO3 P 20.000 motx 106 g = 0.9339. - Mass of wall in sample= 4.980g-0.933g=4.0479 % Composition No. (0₂=0.9338 x 100% = 18.73% 4.9808 % composition of Nad = 4,0478 x 100% = 81.27% 4.9808