Introduction to Chemistry CHM 090 Fall 2017 Assignment 4 Name: Date: 11/29/2017

Question

Introduction to Chemistry CHM 090 Fall 2017 Assignment 4 Name: Date: 11/29/2017 1. Amixture of 150m!ofvinegar acetic acid15%) and 75 mg of sodium bicarbonate is mixed inside a closed 16 fluid ounces container at 750 mmHg of pressure and 52-F. As the compounds mix with each other the chemical reaction shown here takes place. a. Using the chemical reaction provided calculate the amount of carbon dioxide that should be released (theoretical yield). Indicate the limiting reactant in this reaction (using the given values). b. Calculate the pressure of carbon dioxide once the reaction is completed (remember to account for the volume of the solution in your calculations) Calculate the total volume inside of the container (remember Dalton's law of partial pressures). Calculate the total amount of gas particles in the container. You don't have to specify the gases included in that mixture. c. d. Calculate the molar concentration (molarity or M) of sodium acetate (NaC H.O) once the reaction is completed. Draw a graph relating pressure to temperature. Use the measurements from part b as your initial values, then calculate pressure once the temperature is changed to 40 °C, 115 °E, 83 °C, and 381 K List the molecular weight and percent composition for all the chemical compounds listed in the reaction. e. f. List the conversions for all the values that are not given in units of the proper units (temperature in kelvin, mass in grams, volume in liters, etc). CHM 090

Explanation / Answer

mol of NaHCO3 = mass/MW = (75*10^-3)/(84.007 ) = 0.0008927

mol of acid = mass/MW = (D*V)/MW = (x*D*V)/MW = (0.15*150*1)/60 = 0.375 mol of acetic acid

NaHCO3 limtis reaction

mol of CO2expected = 0.0008927 mol of CO2

mass = mol*Mw = 0.0008927*44 = 0.039278 g or 39.27 mg

c)

total gas particles in container...

mol of CO2 --> from ideal gas

PV = nRT

V = 16 oz = 473 mL = 0.473 L

T = 52°F = 11.11

n = PV/(RT) = (750-9.8)(0.473 )/(62.4*(11+273))

n = 0.0197563mol

1 mol = 6.022*10'^23 molecules

0.0197563mol --> (0.0197563)(6.022*10^23) = 1.189*10^22 molecules of CO2

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