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)