Question

Data Table: Trial 1 Trial 2 1. Mass of evaporating dish with watch glass 84.06 g 84.06 g 2. Mass of sodium bicarbonate 2.16 g 2.36 g 85.40 g 85.59 g 85.40 g 85.49 g 3. Mass of dish/watch glass and residue after 1" heating 4. Mass of dish/watch glass and residue after 2 heating 5. Mass of dish/watch glass and residue after 3"| heating 6. Mass of residue (NaCl) (This is the experimental yield of NaCl) 85.49 g Calculations: A. For each trial, calculate the experimental yield (in grams) of NaCl (row 6 in table) by using data from the table. Recall, once the water was completely removed by heating, the dish only contains Naci. B. For each trial, calculate the theoretical yield (in grams) of sodium chloride that could have been produced. This is done with a stoichiometry calculation starting with the mass of sodium bicarbonate used in each trial (row 2 in table). C. For each trial, calculate the % yield of NaCl using the values from the calculations above and the equation for % yield. D. Unrelated to the experimental data, answer the following question: Suppose 1.07 g of water was produced in the same reaction, calculate the mass of sodium bicarbonate that reacted. HCl (aq) + NaHCO3(s) → NaCl (aq) + CO2 (g) + H20 (1)

Answer 1

Answers (Mo). Mass residue (NaCl)=( Mass,o dish watch glass ond) (A.). Mass residue (NaCl)=( residue after and heating! Trial 1 - Mass I evaporating dish) .with watch glass , - oi Massg residueſ Naci) = Experimental yeild... 3 =(85:40 – 84.06) 9 00.g. i = [ 1.367](briat-13 Mass of residue (NaCl) Trial-2= (row_ 5) – (row. 1). = 85.49 – 84.06 :: 1.439./(Trial-2). (B) Hatteoz each mole NaHCO3 will give 1 mole Nau. ... Molar ratio, NaHCO3 : Naci = 1:1. given mass of NaHCO3 = 2.16 9. (troial- 1). Molar mass g NaHCO3 = 23+1+12+(1683) = 24+12 +48 = 849 mole of Nah coz = given mass/molar moss .- 2016/84 = 0.026 mol. Nahcoz. :: 0.026 mole NaHC03 gires. 0.026 mo. Nall. .; 0.026 mol NaCl = 0.026 mol x Molar mass of Nace = 0.026 xd58.5 grams = 1.521.9 ::-; Theoretical yied (Trial-2) -71.52191 1.5210

138 ģ. For trial -2. giver mass of NaHCO3 = 2.369 moles of NaHCO3 = given mass Molar mass = 2.36/84 = 0.028 g moles. oi moles y Nacl produced = 0.028 moles.. Mass g 0.0 28 mot Nacl = 0.028% 58.5. : 1.638 grams.. • Theoretical yeild (trial -2) = 11.638 97. (e) Trial- 1. Trial 2 %yield = Exp-yield x100 %yield- Exp: yield Theoreticał* 100 yield Theoretical yield 2.36 X100 1.521 4 = | 89.41%) = 1.43 X100. 1.638 = 187-30%. 18.9/mot = 0.039 = 0.06 |(D)given reaction: Hel (496) + NaHCO3(s). : → Naul (aq) + CO2(g) + H2009). Molar ratio NaHCO3 • H20. = 1:1. 1.079 water produced = 1.07&_ = 0.059 < 0.06 mol water produce .; 0.06 molé. NaHCO3 must be used up. Molar mass NaHCO3 = 849/mol. Mass of 0.06 mote NaHCO3 = 0:06 X 84 g/mol = 5.50049 ; 15.049] Nay coz is reacted. ::.: Man