90 g What is the molecular formula of the compound 10 points m V PM RT m690g

90 g what is the molecular formula of the compound 10

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weighs 6.90 g. What is the molecular formula of the compound? (10 points) m V = PM RT m=6.90g P=750mmHg x 1 atm 760 mmHg = 0.987 atm T= 393.15 ° K R=gas constant= 0.0821 L atm K mol V=1.00L M= mRT VP = 6.90 g ( 0.0821 L atm K mol ) 393.15 ° K 1.00 L ( 0.987 atm ) = 225.65g/mol C=12.01g/mol H=1.008g/mol O=16.00g/mol C=64.9% x 225.65g/mol = 146.45g/mol / 12.01g/mol=12.19 H=13.5% x 225.65g/mol = 30.46g/mol / 1.008g/mol= 30.22 O=21.6% x 225.65g/mol = 48.74g/mol / 16.00g/mol= 3.05 C=12.19/3.05 = 3.99 H= 30.22/3.05 = 9.91 O=3.05/3.05 = 1 C 4 H 10 O 9. What is the mass of the solid NH 4 Cl formed when 73.0 g of NH 3 ( g ) are mixed with an equal mass of gaseous HCl? What is the volume and identity of the gas remaining, measured at 14.0°C and 752 mmHg? (8 points) NH 3 ( g ) + HCl( g ) → NH 4 Cl NH 3 = 17.034g/mol HCl = 36.458g/mol NH 4 Cl = 53.492g/mol 73gNH 3 x 1 mol NH 3 17.034 g NH 3 x 1 mol NH 4 Cl 1 mol NH 3 x 53.492 gNH 4 Cl 1 mol NH 4 Cl = 229.242 g NH 4 Cl
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73gHCl x 1 mol HCl 36.458 g HCl x 1 mol NH 4 Cl 1 mol NH 3 x 53.492 gNH 4 Cl 1 mol NH 4 Cl = 107.107g NH 4 Cl 107.11g NH 4 Cl formed 229.42g NH 4 Cl – 107.107g NH 4 Cl = 122.313g NH 4 Cl 122.313g NH 4 Cl x 1 mol NH 4 Cl 53.492 gNH 4 Cl x 1 mol NH 3 1 mol NH 4 Cl x 17.034 g NH 3 1 mol NH 3 = 38.949 g of NH 3 remaining m V = PM RT m= 38.949 g P=752mmHg/760mmHg = 0.989atm M=17.034g/mol T=287.15°K R=gas constant= 0.0821 L atm K mol V= mRT PM = 38.949 g ( 0.0821 L atm K mol ) 287.15 ° K 0.989 atm ( 17.034 g mol ) = 54.5L of NH 3 remaining (Reference: Chang 5.59) 10. A mixture of gases contains 0.31 mol CH 4 , 0.25 mol C 2 H 6, and 0.29 mol C 3 H 8 . The total pressure is 1.50 atm. Calculate the partial pressures of the gases. (8 points) X i = ¿ nt P=X i P t P t = P CH4 + P C2H6 + P C3H8 X CH4 = 0.31 0.85 = 0.37 X C2H6 = 0.25 0.85 = 0.29 X C3H8 = 0.29 0.85 = 0.34 P CH4 = 0.37 x 1.50 atm = 0.56 atm CH
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