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A sample of pure sodium carbonate with a mass of 0.452 g was dissolved in water and reacted

A sample of pure sodium carbonate with a mass of 0.452 g was dissolved in water and reacted with 42.5 ml of 0.250 M hydrochloric acid. Write the balanced equation and determine the percent yield when the volume of the gas product collected at 0 o C and 760. mm Hg was 89.7 ml.

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A sample of pure sodium carbonate with a mass of 0.452 g was dissolved in water and reacted

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  1. A sample of pure sodium carbonate with a mass of 0.452 g was dissolved in water and reacted with 42.5 ml of 0.250 M hydrochloric acid. Write the balanced equation and determine the percent yield when the volume of the gas product collected at 0 oC and 760. mm Hg was 89.7 ml.

  2. Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2 (g) 0.452 g 42.5 ml 0.250 M 1 CO2 = 0.00426 mol CO2 1 Na2CO3 0.452 g 1 mol = .00426 mol 106.0 g 0.0425 L x 0.250 M = 0.0106 mol 1 CO2 = 0.00531 mol CO2 2 HCl 0 oC & 760. mm Hg = STP @ STP : 22.4 L mol x 0.00426 mol CO2 = 0.0954 L

  3. or use PV = nRT R = 0.0821 L atm/mol K 0 oC = 273 K 760. mm Hg = 1 Atm n = 0.00426 mol CO2 V = (0.00426) (0.0821) (273)= 0.0955 L = 95.5 mL 1.00 89.7 ml = 93.9% 95.5 mL

  4. Determine the percent yield when 432 mL of gas is obtained at 27oC and 700. torr by reacting 5.00 g of zinc metal with 100. mL of 0.350 MHCl.

  5. Zn (s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) 5.00 g 100. ml 0.350 M 5.00 g 1 mol = .0765 mol 65.4 g 1 H2 = 0.0765 mol H2 1 Zn 0.100 L x 0.350 M = 0.0350 mol 1 H2 = 0.0175 mol H2 2 HCl 0 oC & 760. mm Hg = STP @ STP : 22.4 L mol x 0.0175 mol H2 = 0.392 L

  6. 27+ 273 = 300. K 700./760. = 0.921 atm 1 atm(0.392 L) = (0.921 atm)V2 273 K 300. K Using P1V1 = P2V2 : T1 T2 V2 = 0.468 L or use PV = nRT n = 0.0175 mol H2 V = (0.0175) (0.0821) (300) = 0.468 L = 468 mL 0.921 432 ml = 92.3% 468 mL

  7. Honors Chemistry - Equations & Constants NA = 6.02 x 1023 PV = nRT R = 0.0821 L atm / mol K P1V1 = P2V2 K = oC + 273 1 atm = 760 mmHg T1 T2 STP = 0oC and 1 atm Ptotal= P1 + P2 + P3 + … @ STP 1 mol = 22.4 L q = m ∆T c (Heat = mass x temp change x specific heat) pH = - log[H+] pOH= - log[OH-] pH + pOH= 14

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