Question

how do i figure out the moles of CO2 and NaHCO3 along with the mass of the sample?

Atmospheric pressure, mm Hg 423.66 mmHg 25.0°C Temperature of H,O, °C 238 mm Hg Vapor pressure of H,O at this temperature, mm Hg Pressure of Co, in the system, mm Hg! 599.8 mmHg NaHCO, + HCl-NaCl + CO, + H,O (MM NaHCO, = 84.01 g/m Trial 2 3 Mass of sample, g 1 Vinitiae mL 2 1.70 mL 1.03mL 1.02mL V anal ml 14.03mL 8.03ml 15.07ml V mL (V -V +0.012332 -0.0074 -0.01405L - 12.33mL 7 ml - 14.05 mL Moles of CO2 0.0118 0.0349 0.0139 Moles of NaHCO, Mass of NaHCO, in sample,g Mass % of NaHCO, in sample, % Average mass % NaHCO, in Alka-Seltzer Show your calculations for Trial I on the next page, and submit them with your Data Sh

Answer 1

Reaction is NaHCO + HCl → Nacl + 10+ c & from the above equation imol Natco liberates Tmol co ie (22.4L) at STP conditions. Volume of co produced from the sample at 256 and lab pressure should be converted to Vo, by the equation Povo - PV : P = Pen-V.p of Ho STP" To Talab 2 (623.6-23.8) mm Hg = 599.8 Trial mass of sample 0.0719 0.0349 0.0739 V cm) (V, -V:) 24.03–170)m) (8.03 - 1-03) | = (15-07-1-02)ml = 12.33 ml. 7ml. = 14.05 ml. ann 2 5.06m1 10.16 ml. 2102ml. -4 2.26x 10' 4.56x10 18.91m) -3.98x10' Vo = Pv. To (399.8 mm Hg) (2.33 mi)(273) K POT (760mm Hg) (298k) = 8.91ml Moles of CO = V celo 22,400 ml 22,400ml moles of NaHCO₃ = 3.98 x 104 moles & Co Mass Naltco = Moles x (84.01 g/mol) O molar mass = 3.34x10²g 2-26x104 4.566104 (3-98x10 "mol) x 1.90x10 3.83x10²g. Mass / Nahco = Mass of NaHg x100 3.83x10 9 x100 3 mass of sample 1.96810g x 100 13.34x10 9 x100 0.0739 10.0349 0.0719 = 55.9.1 = 52.54 = 47.04 / - (47.04 +55.9 + 52.5) % - 51.8 % = ! Average mass of NaHCg