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We can then determine the percent by mass of CaCO3in the 3.00 g sample. The balanced equation is: CaCO3(s) +2HCl(aq) ⎯⎯→CO2(g) +CaCl2(aq) +H2O(l) The moles of CO2produced can be calculated using the ideal gas equation. 22COCO=PVnRT2CO1 atm792 mmHg(0.656 L)760 mmHgL atm0.0821(20 + 273 K)mol K×==⋅⋅222.8410mol COn−×The balanced equation shows a 1:1 mole ratio between CO2and CaCO3. Therefore, 2.84 ×10−2mole of CaCO3must have reacted. 233333100.1 g CaCO? g CaCOreacted(2.8410mol CaCO )2.84 g CaCO1 mol CaCO−=××=The percent by mass of the CaCO3sample is: 2.84 g100%3.00 g=×=3% CaCO94.7%Assumption:The impurity (or impurities) must not react with HCl to produce CO2gas. 5.61 The balanced equation is: H2(g) +Cl2(g) ⎯⎯→2HCl(g) At STP, 1 mole of an ideal gas occupies a volume of 22.41 L. We can use this as a conversion factor to find the moles of H2reacted. Then, we can calculate the mass of HCl produced. 222221 mol H? mol Hreacted5.6 L H0.25 mol H22.41 L H=×=
CHAPTER 5: GASES 149The mass of HCl produced is: 222 mol HCl36.46 g HCl0.25 mol H1 mol H1 mol HCl=××=? g HCl18 g HCl5.62 The balanced equation is: C2H5OH(l) +3O2(g) ⎯⎯→2CO2(g) +3H2O(l) The moles of O2needed to react with 227 g ethanol are: 25225225251 mol C H OH3 mol O227 g C H OH14.8 mol O46.07 g C H OH1 mol C H OH××=14.8 moles of O2correspond to a volume of: 222O2O2L atm(14.8 mol O ) 0.0821(35273 K)mol K3.6010L O1 atm790 mmHg760 mmHg⋅+⋅===××nRTVPSince air is 21.0 percent O2by volume, we can write: 22O222100% air100% air(3.6010L O )21% O21% O==×=3air1.7110 L airVV×5.63 (a)34 L NH12.8 L NO4 L NO×=312.8 L NH25 L O12.8 L NO4 L NO×=216.0 L O(b)38221 L C H8.96 L H7 L H×=381.28 L C H2223 L H O8.96 L H7 L H×=23.84 L H O5.64 The balanced equation is: FeS + 2HCl H2S + FeCl2. From the moles of H2S produced, we can calculate the mass of FeS in the 4.00 g sample. The mass percent purity can then be calculated. 2H S21 atm782 mmHg(0.896 L)760 mmHg0.0391 mol H SL atm0.0821(287 K)mol K×===⋅⋅PVnRTThe mass of FeS in the 4.00 g sample is: 221 mol FeS87.92 g FeS0.0391 mol H S3.44 g FeS1 mol H S1 mol FeS××=CHAPTER 5: GASES 150The mass percent purity is: 3.44 g100%4.00 g=×=mass percent purity86.0%5.67 First, we calculate the mole fraction of each component of the mixture. Then, we can calculate the partial pressure of each component using the equation, Pi=ΧiPT. The number of moles of the combined gases is: 42638CHC HC H0.31 mol0.25 mol0.29 mol0.85 mol=++=++=nnnn42638CHC HC H0.31 mol0.25 mol0.29 mol0.360.290.340.85 mol0.85 mol0.85 mol======ΧΧΧThe partial pressures are: 4CHtotal0.361.50 atm=×=×=4CH0.54 atmPPΧ26C Htotal0.291.50 atm=×=×=26C H0.44 atmPPΧ38C Htotal0.341.50 atm=×=×=38C H0.51 atmPPΧ5.68Dalton's law states that the total pressure of the mixture is the sum of the partial pressures.