Question

Data 50.00 g Mass of beaker: Mass of beaker and sample (NaHCO3): Mass of sample (NaHCO3): s1.00 g 49.57 g 49.57 g Mass of beaker and residue first heating: Mass of beaker and residue second heating: Mass of beaker and residue third heating (if required): Mass of residue: g 0.43 g mL Volume of concentrated HCl added: Calculations Mass of 1 mole of NaHCO3 Number of moles of NaHCO3 added: 84.007 0.0116 g/mole moles 1. M Concentration of HCl added: Number of moles of HCI moles Include all state Write a balanced chemical equation for this chemical reaction. designations for both reactants and products. NaHCO₃ + Hel Nach + H₂O + CO₂ (s) ISI How many moles of HCl are required to react completely with the NaHCO3 you have measured out? 0.0119 moles

which reactant is limiting?
which reactant is in excess?
actual yield of NaCl, show the calculation
percent yield of NaCl, show calculation

Answer 1

Ans. Complete answer is given in picture ,for any doubt feel free to ask in the comment box,give it a Thumbs up Thank-you!

Ans males of Natica 0.0119 mal 0.01 10 mal is males of Hel = Nath co the - Nall tho + Co (S) d) (g) (9) O. IJ I mal o.119 mal As the number of males of Natco and Hee is same similarly stoichiometric coefficient of Natco a Ha same (1) Therefore there is no linting reagent in this creation : There is no reactant in excess. Actual gilld of Nace= 0.43 Imal of Nanco give smal of Nach a. ol19 mel of Nam co give o. osa. of 19 mal of Nace mass of Nace = male x malar mass = 0.0319nbod * 58.49gmat! 2.69549 Theoretical yield of Nace - 0-69549 Pincent yield of Nall-Actual indd x 100% Theoretical freld -J 0.43 0.6954 X100% 61.836 Scanned with CamScanner CS Scanned with CamScanner