Question

2.50 L of N2 at 25 C and 1.04 atm is mixed with 3.00 L of 02 at 25'C and 86.6 mm Hg, and the mixture is allowed to react. How much NO, in grams, is produced? N2(g)+O2(g)-2NO(g) A 8.40 g NO is produced B 4.80 g NO is produced C 0.840 g NO is produced D 0.480 g NO is produced

Answer 1

2.50L Given ! VN2 = TN2 = 25% PN2 = 1.04 atm, 0.114 atm. = 86 86.6 am Po2 = 86.6 mm Hg 760 vo2 = 31. TO2 = 25°6 = 298 K since [pv=nRT] where is the mod L e J/mol. R ser 0.082 = atm. L. molk 8.314 R gas constant = n = PV = nwe = PN2. UNA R- TN2 RT No = = 0.106 mol 1.04% 2.50 0-089 x 298 no2 = Pozivoa = 0.11483 0.0888 298 R. TOR noa = 0.014 mol since mol of oa Inoa) a mot of No 10 N2) so 02 is limiting beagent. 1 mol Od gives a mol No 1g) In given reaction , 2x no2 = 0.028 mol so mol of No = molas mass of no = 30 g/mol so mass of no = 10.028 mol) (30 g/mol) m = 10.8409) option (c)