Lab Report 5.pdf - Name Daniela Tapia Chem 105\/110 Section Lab Partner Experiment Date Mole Ratios and Reaction Stoichiometry Reaction A Sodium

Lab Report 5.pdf - Name Daniela Tapia Chem 105/110 Section...

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Unformatted text preview: Name: Daniela Tapia Chem 105/110 Section: Lab Partner: Experiment Date: 03/20/20 Mole Ratios and Reaction Stoichiometry Reaction A: Sodium Bicarbonate and Hydrochloric Acid Experimental Data (a) Mass of evaporating dish + watch glass 73.28g (b) Mass of evaporating dish + watch glass + sodium bicarbonate 75.77g (c) Mass of sodium bicarbonate used 2.50g (d) Mass of evaporating dish + watch glass + sodium chloride 74.91g (e) Mass of sodium chloride collected (experimental yield) 1.63g Data Analysis 1) Use your data to determine the experimental mole-to-mole ratio between sodium bicarbonate and sodium chloride. Show your work for each step. • Convert the mass of sodium bicarbonate used to moles. ____________ 2.50gNaHCO3 x 1mol NaHCO3 84.01gNaHCO3 • -2 = 2.97 x 10 mol Convert the mass of sodium chloride collected to moles. 1.63gNaCl x___________ 1mol NaCl -2 = 2.79 x 10 mol 58.44gNaCl • Divide both of your results from the preceding two steps by the lower mole value to determine the simplest mole-to-mole ratio between sodium bicarbonate and sodium chloride. 2.97 ____ 2.79 _____ 2.79 = 1 2.79 = 1.06 mole -mole ratio Simplest mole ratio before rounding 1.06 moles NaHCO3 : 1 moles NaCl 1 moles NaCl Simplest whole number mole ratio after rounding 1 moles NaHCO3 : Page 1 of 3 2) Determine your percent yield of sodium chloride in reaction A. Show your work for each step. • • Write the balanced equation for reaction A – the reaction between sodium bicarbonate and hydrochloric acid. 1NaHCO3+1HCl ______ 1NaCl +1CO2 +1H2O (l) (s) (g) (aq) (aq) Using mass-to-mass stoichiometry, calculate the theoretical yield of NaCl for reaction A. Use your initial mass of sodium bicarbonate reactant as a starting point, along with the relevant mole ratio from the balanced equation to perform this calculation. 2.50gNaHCO3 x 1mol NaHCO3 x__________ 1mol NaCl x 58.44gNaCl _____________ ___________ _________ =1.74g 84.00g NaHCO3 1molNaHCO3 1molNaCl • Calculate your percent yield of sodium chloride product. 1.63g x 100% = 93.68% ______ 1.74g Reaction B: Sodium Carbonate and Hydrochloric Acid Experimental Data (a) Mass of evaporating dish + watch glass 33.076g (b) Mass of evaporating dish + watch glass + sodium carbonate 32.501g (c) Mass of sodium carbonate used 0.575g (d) Mass of evaporating dish + watch glass + sodium chloride 33.653g (e) Mass of sodium chloride collected (experimental yield) 0.577g Data Analysis 1) Use your data to determine the experimental mole-to-mole ratio between sodium carbonate and sodium chloride. Show your work for each step. • Convert the mass of sodium carbonate used to moles. -3 0.575gNa2CO3 x 1mol Na2CO3 = 5.43 x 10 mol ___________ 105.99g Na2CO3 • Convert the mass of sodium chloride collected to moles. -3 0.577gNaCl x _________ 1mol NaCl = 9.87 x10 mol 58.44g NaCl Page 2 of 3 • Divide both of your results from the preceding two steps by the lower mole value to determine the simplest mole-to-mole ratio between sodium carbonate and sodium chloride. ____ 9.87 = 1.82 5.43 Simplest mole ratio before rounding 1.82 moles Na2CO3 : 1 moles NaCl Simplest whole number mole ratio after rounding 1 moles Na2CO3 : 1 moles NaCl 2) Determine your percent yield of sodium chloride in reaction B. Show your work for each step. • Write the balanced equation for reaction B – the reaction between sodium carbonate and hydrochloric acid. 1Na2CO3 +2HCl ——— 2NaCl + 1C02 +1H2O • Using mass-to-mass stoichiometry, calculate the theoretical yield of NaCl for reaction B. Use your initial mass of sodium carbonate reactant as a starting point, along with the relevant mole ratio from the balanced equation to perform this calculation. 0.575gNa2CO3 x ___________ 1mol NaCO3 x 1mol NaCl x 58.44gNaCl _________ ________ = 0.317g 105.99g NaCO3 1mol NaCO3 1mol NaCl • Calculate your percent yield of sodium chloride product. _____ 0.577 x 100% = 182.02% 0.317 3) Is your percent yield here for reaction B greater than or less than 100%? Give one possible source of error that could explain the percent yield you obtained. This is greater than 100%, probably this was because there are too much HCL added to the solution. Page 3 of 3 ...
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