We can then determine the percent by mass of CaCO 3 in the 300 g sample The

# We can then determine the percent by mass of caco 3

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We can then determine the percent by mass of CaCO 3 in the 3.00 g sample. The balanced equation is: CaCO 3 ( s ) + 2HCl( aq ) ⎯⎯→ CO 2 ( g ) + CaCl 2 ( aq ) + H 2 O( l ) The moles of CO 2 produced can be calculated using the ideal gas equation. 2 2 CO CO = PV n RT 2 CO 1 atm 792 mmHg (0.656 L) 760 mmHg L atm 0.0821 (20 + 273 K) mol K × = = 2 2 2.84 10 mol CO n × The balanced equation shows a 1:1 mole ratio between CO 2 and CaCO 3 . Therefore, 2.84 × 10 2 mole of CaCO 3 must have reacted. 2 3 3 3 3 3 100.1 g CaCO ? g CaCO reacted (2.84 10 mol CaCO ) 2.84 g CaCO 1 mol CaCO = × × = The percent by mass of the CaCO 3 sample is: 2.84 g 100% 3.00 g = × = 3 % CaCO 94.7% Assumption: The impurity (or impurities) must not react with HCl to produce CO 2 gas. 5.61 The balanced equation is: H 2 ( g ) + Cl 2 ( 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 H 2 reacted. Then, we can calculate the mass of HCl produced. 2 2 2 2 2 1 mol H ? mol H reacted 5.6 L H 0.25 mol H 22.41 L H = × =
CHAPTER 5: GASES 149 The mass of HCl produced is: 2 2 2 mol HCl 36.46 g HCl 0.25 mol H 1 mol H 1 mol HCl = × × = ? g HCl 18 g HCl 5.62 The balanced equation is: C 2 H 5 OH( l ) + 3O 2 ( g ) ⎯⎯→ 2CO 2 ( g ) + 3H 2 O( l ) The moles of O 2 needed to react with 227 g ethanol are: 2 5 2 2 5 2 2 5 2 5 1 mol C H OH 3 mol O 227 g C H OH 14.8 mol O 46.07 g C H OH 1 mol C H OH × × = 14.8 moles of O 2 correspond to a volume of: 2 2 2 O 2 O 2 L atm (14.8 mol O ) 0.0821 (35 273 K) mol K 3.60 10 L O 1 atm 790 mmHg 760 mmHg + = = = × × n RT V P Since air is 21.0 percent O 2 by volume, we can write: 2 2 O 2 2 2 100% air 100% air (3.60 10 L O ) 21% O 21% O = = × = 3 air 1.71 10 L air V V × 5.63 (a) 3 4 L NH 12.8 L NO 4 L NO × = 3 12.8 L NH 2 5 L O 12.8 L NO 4 L NO × = 2 16.0 L O (b) 3 8 2 2 1 L C H 8.96 L H 7 L H × = 3 8 1.28 L C H 2 2 2 3 L H O 8.96 L H 7 L H × = 2 3.84 L H O 5.64 The balanced equation is: FeS + 2HCl H 2 S + FeCl 2 . From the moles of H 2 S produced, we can calculate the mass of FeS in the 4.00 g sample. The mass percent purity can then be calculated. 2 H S 2 1 atm 782 mmHg (0.896 L) 760 mmHg 0.0391 mol H S L atm 0.0821 (287 K) mol K × = = = PV n RT The mass of FeS in the 4.00 g sample is: 2 2 1 mol FeS 87.92 g FeS 0.0391 mol H S 3.44 g FeS 1 mol H S 1 mol FeS × × = CHAPTER 5: GASES 150 The mass percent purity is: 3.44 g 100% 4.00 g = × = mass percent purity 86.0% 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, P i = Χ i P T . The number of moles of the combined gases is: 4 2 6 3 8 CH C H C H 0.31 mol 0.25 mol 0.29 mol 0.85 mol = + + = + + = n n n n 4 2 6 3 8 CH C H C H 0.31 mol 0.25 mol 0.29 mol 0.36 0.29 0.34 0.85 mol 0.85 mol 0.85 mol = = = = = = Χ Χ Χ The partial pressures are: 4 CH total 0.36 1.50 atm = × = × = 4 CH 0.54 atm P P Χ 2 6 C H total 0.29 1.50 atm = × = × = 2 6 C H 0.44 atm P P Χ 3 8 C H total 0.34 1.50 atm = × = × = 3 8 C H 0.51 atm P P Χ 5.68 Dalton's law states that the total pressure of the mixture is the sum of the partial pressures.