A gaseous mixture consists of 6.59 g of N2, 4.07 g of O2, and 2.95 g of He. What volume does this mixture occupy at 21°C and 1.35 atm pressure?
2007-04-30 20:12:50 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
6.59 / 28 = 0.235 mole N2
4.07 / 32 = 0.127 mole O2
2.95 / 4 = 0.737 mole He
Total moles = 1.099
T = 21 + 273 =294 K
V = nRT / p = 1.099 x 0.0821 x 294 / 1.35 = 19.6 L
2007-04-30 20:20:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Moles of N2 = 6.59/28 = 0.235
Moles of O2 = 0.127
Moles of He = 0.738
Thus vol = nRT / P = 19.67 l
2007-05-01 03:24:34 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋
6.59g of N2 = 6.52/28 =0.233mole of N2
4.07g of O2= 4.07/32=0.127 mole of O2
2.95g of He = 2.95/4=0.738mole of He
so total number of moles of gas is
n=0.233+0.127+0.738=1.098 moles of gas
and pV = nRT
V = nRT/P
in MKSA
V= 1.098*8.31*(273+21)/(1.35*101325)=0.0196m^3 = 19.6L
2007-05-01 03:22:11 · answer #3 · answered by maussy 7 · 0⤊ 0⤋