Question

Section Date OBA DATA SHEET: EXPERIMENT 15 1. Mass of beaker Mass of beaker and NaHCO, 3. Mass of NaHCO3 8.0 E m m Volume of 3.0 M HCl required to react with NaHCO3 (Show set-up) lag etcos Imotno ex lorot tcl x 1000 +1+2.01 +10x Sin = e lohatos mo _ml 5. Actual volume of 3.0 M HCl used (after 15" heating) 65.53 26.48 6. Mass of beaker and NaCl (after 2nd heating) (after 3d heating, if needed) 7. Actual mass of NaCl produced in reaction 8. Mass of NaCl predicted by stoichiometry (Show set-up) Percent yield of NaCl (Show set-up) 145

Answer 1

Percent yield can be more than 100% if the product is not dried properly or it contains some impurities like unreacted reactant.

Reaction Involved : NaHCO3(s) + H cecast Nacest H₂O + CO₂(g) Mass of beaker and Nall = 65.48 g. I Mass of beaker = 62.909 7. Actual mass of Nall produced = 65-489-62.909 = 2058 g 81 mol NaHCO3 react with Imol Hil to form Imol Nall. Mass NaHCO3 = 10995g molar mass NaHW ₃ = 84.00791 Moles Nah co2 = mass 10995g - Molar mass - 84007 g/mol =0.0237 moles Moles HCl = molarity X Volume = 3.0M X 128 ml x L = 0.042 moles. 1000 ml Moles Naho are lesser than Moles He. So NaHCO3 is the limiting reagent. 0.0237 mol NaHCO3 will produce 0.0237 mol Nace. Mass nace = moles X Molar mass -0.0237 mol x 58.44 g/mol = 1.38g 9. Percent yield - Experimental yield 100 theoretical = 2.58 g x 100 = 186.9% 1.38g