Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: afchick7689 on September 04, 2005, 02:23:34 PM
-
Heres the problem, my work, and my answer--
if someone could please check over this, i think i must have made an error somewhere because the answer seems very large...
What would be the volume occupied by 1.0 kg of carbon monoxide at 700 °C and 0.1 atm?
PV= nRT
0.1 atm * V = n * 0.0821 * (700 + 273.15)
To find n --
1.0 kg CO | 1000 g CO = 1000 g CO
| 1 kg CO
molar mass of CO = (12.01) + (16.00) = 28.01g
1000 g CO | 1 mol CO = 35.702 mol CO
| 28.01 g CO
...SO:
0.1 atm * V = 35.702 mol CO*0.0821*(700 + 273.15)
.01 V = 2852.43
V= 285243.32 L (W/ SIG FIGS: 300000 L)
-
The second-to-last line should read:
0.1 V = 2852.43
Which is why your answer is off by a factor of 10.
-
thanks!