Question

5.(16 pts, 8min) Do either (a) or (b) (a) A sample of Ar(g) having a mass of 0.0166 g is collected over water. If the temperature and total pressure of the system are 48.0°C and 676 mmHg respectively, what is the volume of the gas expressed in liters? Data: is 83.71 mmHg. R= 0.0821 L atm/(K mol), 1 atm 760 mmHg Atomic mass of Ar-39.95 and the vapor pressure of water at 48.0°C (b) Consider the following reaction: 2 NAOH(aq)+2 Al(s) +6 H2O()- Calculate the volume, in Liters, of H2(g) at 1.22 atm and 26.3 °C produced from 4.75 g of Al(s) assuming we have a sufficient amount of NaOH and H2O. Data: Atomic masses Al = 26.98, H 1.008, Na- 22.99, O- 16.00, R = 0.0821L atm/(K mol) 2 NaAl(OH)4(aq) + 3 H2(g)

Answer 1

ways of AG (W) = 0.01669 atomie mass (MW) = 39.95g/mol .: mots q Ar (0) - Mo 39.95g/ml i mols & Ar (n) = -0.01669 T=48°c =3211 n=4-155x10tmoly R = 0.0821 halmolek mol Pon= 676 mmittg : PH2org) = 83.71 mmHg Par = PGas - PH2O = 676-83-7) Par = 592.29 mmity P = 0.779 atro PV = nRT - RT 4/55XD x 0-0821 x 32) 0.779 = 0.014 L volume of Ar=0.0142 Al +2 NaOH + H2O -Q NOAI (047 +3Hqc9) amoly ☺ 3mol mass of Al =4.75g ; Atomie mass of Al = 26-98 .: No. of mots of Al = 4.259 = 0.176 mol p=1-22 atm : v= ? T = 26.34 = 299 From Ras amos al produces 3 mol H2 gas .: 0:176 mol H 2 P = 0476 x 2 = 0.2.64 mol volume of H₂ = 26 mol x 0:0821 x 299.4k - 532L 1:22 al to