Question

Table Salt from Baking Soda Name: Date: Partner: DATA: Mass of beaker 95.85. Mass of beaker and sample (NaHCO3) 17.45. Mass of sample (NaHCO3) Mass of beaker and residue (NaCl) 1" Heating... :26 99, Mass of beaker and residue (NaCl) 2 Heating.. 26.85, Mass of beaker and residue (NaCl) 3* Heating... 26.959 Mass of beaker and residue (NaCl) best g Mass of residue (NaCl) 9 Volume of conc. HCI added ml CALCULATIONS: g/mol Molar mass of NaHCO3 Number of moles of NaHCO3w added Show calculations below: mol Concentration of HCI (aq) M Number of moles of HCI added mol Show calculations below: Write a balanced chemical equation for this chemical reaction. Include state designations for all reactants and products

Answer 1

(1) Mass of Beaker = 25.85 g Mass of Beaker & Sample 1 NaHCO3) = 27.459 Mass of Sample (NaHCO3) = 27.459-25.859 1.6 g (NaHCO₃). => Mass of Beaker and residue (Nad) best = 26.85. => Because heating Constant. =) Mass of residue (Nach) = 26.85g=25.859 mass remains after se cound 1g ( Nad) as residue. removed Which of NaHCO3 S => Due to Heating water formed during the reaction. Volume of conc. Hel added 4.0mL Calcuations: Molar mass = 84.007 g/mol Number of moles of Natcoz added Given mass of Nalco, Molar mass Number of miles = libg = 10.01904 modes of NaHCO3 84.007 Concentration of Helag, = 12m volume of Helags added 4.50mL we Know that = Number of moles of Hel added volume of solution in L 1 2 moles X Number of moles of He added 4.58103k => Number o of Hel added 12 x 4.5 x 10 moles 10.054 Now Converting to grams we have Number of moles Given mass of Hal added Molar mass of He 0.054 mosh, x 36.5g/mté 11.971g of Hel added ... 18 Moles -3 moles moles

The Balanced Chemical equation is given as fallows NaHCO3 + Hel → Nach + H₂O + CO2 so 993 (99) 92 Sto Chiometry reacts with 1 mole to 1 mole 58.5g of Nach g 84.007 58.5*1-6 مه 1.liug lillig Assuming Hel ii) Assuming NaHCO3 Limiting Reagent. =) from the of Chemical equation We have 1 mole of Nallcoa of Hel give of Nace. => 84.0079 of Nall coz = lg of Nallco z = 5815 .. 1.6g of NaHCO3 84.007 Thus 1.6 g of Natl Coz produces Nad = be limiting Reagent. = 36.5g of Hel - 58.5g of Nach g of 58.5g => 1.971g of Hel -g=13-15agl Then from 1.9719 of H4, 3.159g of Nad is produced, Thus Nancoz is limiting Reagent, because it first This Her is Eaces reagent Actual yield of Nad = 19 theoretical yield of Nad = 1- 114g Actual giend x 100 19 x 100 = 189.76% Theoretical yield Since He is reagent Some amount of it used. in the product. Hun fume hood Ha 36.5 58.5 X 1.971 36.5 i consumed in the reaction. % yield 1.114g Excess is present still