Open with The Extra Problems for CHEM 111 A 19.06 g mixture containing MgCO, ()
ID: 567929 • Letter: O
Question
Open with The Extra Problems for CHEM 111 A 19.06 g mixture containing MgCO, () is heated until the MgCO, decomposes to MgO and CO. The carbon dioxide is collected over water in a container al containing Ne and water vapor. The total volume is 655.0 ml at 25.0 °C. The atmospheric pressure is 67.7 cm and the vapor pressure of water at 25.0 °C is 23.8 torr The Ne is removed and placed into another 1.35 L container. The Ne exerts a pressure of 155 mm Hg at 25.0°C. (The molar mass of MgCO, is 84.31) 1) MgCO3 (s) MgO(s) + CO2 (g) Calculate the % MgCO, in the mixture. 21 A studantExplanation / Answer
first, moles of neon need to be calculated using gas law equation
from gas law, moles of neon = PV/RT, P = 155 mm Hg, in terms of atmospheres, P= 155/760 atm =0.203 atm
V= volume in L= 1.35L, T= 25+273= 298K, R= gas constant =0.0821 L.atm/mole.K
n= moles of neon = 0.203* 1.35/(0.0821*298)=0.011 moles
moles of CO2 and Neon can be calculated fom partial pressures of CO2 and Neon. This can be obtained by substracting the total pressure- saturation vapor pressure of water=
76 cm of Hg is 1 atm and 67.7 cm is 67.7/76 atm =0.89 atm, 1 atm =760 Torr
0.89 atm =0.89*760 torr = 676.4 Torr, vapor pressure of water= 23.8 Torr
partial pressure of dry gas,P = 676.4-23.8=652.6 Torr, in terms of atm, Partial pressure of CO2+Ne= 652.6/760 atm=0.86 atm, V= 655ml= 655/1000L=0.655 L, T= 25 deg.c= 25+273= 298K, no of moles of CO2+Ne= PV/RT= 0.86* 0.655/(0.0821*298)= 0.023 moles
hence moles of CO2= moles of CO2+Ne - moles of Ne= 0.023-0.011=0.012
from the reaction, MgCO3(s) ------->MgO(s)+CO2(g)
1 mole of CO2 requires 1 mole of MgCO3.
0.012 moles of CO2 requires 0.012 moles of MgCO3, molar mass of MgCO3= 84.31g/mole
mass of MgCO3= moles* molar mass =0.012*84.31 =1 gm
mass % of MgCO3= (1/19.06)*100 = 5.25%
mass of Mixture = 19.06 gm
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