Determine how many liters 8.47g of carbon dioxide gas would occupy at the following conditions.

Question

STP

140°C and 5.00 atm

239K and 109 laps

Answer 1

Molar mass carbon dioxide (CO2) = 44.0 g / mole

8.47 g CO2 x 1 mol / 44.0 g = 0.1925 moles CO2

At STP, 1 mole of any ideal gas occupies 22.4 liters, thus

@STP: 0.1925 mols CO2 x 22.4 L / 1 mol = 4.31 L

At 140ºC and 5.00 atm, we will use the ideal gas law:

PV = nRT and V = nRT/P

V = (0.1925 mol)(0.0821 Latm/Kmol)(413K) / 5.00 atm

V = 1.31 L

At 239K and 109 laps(?) .. never heard of laps units, so will assume you meant kPa (kilopascals)

109 kPa x 1 atm / 101.325 kPa = 1.076 atm

PV = nRT and V = nRT/P

V = (0.1925 mol)(0.0821 Latm/Kmol)(239K) / 1.076 atm

V = 3.51 L