What pressure will 14.0 g of CO exert in a 3.5 L container at 75°C?

moles CO = 14.0 g /28.0 g/mol= 0.500
T = 75 + 273 =348 K
p = nRT/ V = 0.500 x 0.08206 x 348 / 3.5 L=4.08 L

14 g of Co = 14/28 = 0.5 mole. so n=0.5, R=0.082atm L mol-1 K-1, T= 273+75 =348K, V=3.5 L
so applying gas equation, PV=nRT,
P= 0.5X0.082 X348/3.5 =4.0765 atm.

If you are given
V = 3.5 L

T = 75 degrees Celsius = 348 K ( from adding 273 and 75)

R = 0.082057 L atm / K mol (gas constant)

n = number of moles ( obtained from the given mass of CO )

14.0g CO ( 1 molCO / 28gCO (molar mass) ) = .5 mol CO

P = looking for pressure (atm)
then just plug in values in the ideal gas law equation:
PV= nRT

where P = nRT / V
P = (.5molCO) (0.082057 L atm/K mol) (348K) / (3.5 L) = 4.07 atm
Pressure of CO is 4.07 atm