What is the volume occupied by 2.00 moles of ar (g) at standard temperature and pressure (stp)?

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In order to solve this problem we would use the Ideal Gas Law formula #PV=nRT#

#P =# Pressure in #atm#
#V=# Volume in #L#
#n=# moles
#R=# Ideal Gas Law Constant
#T=# Temp in #K#

#STP# is Standard Temperature and Pressure which has values of
#1 atm# and #273K#

#2.34g CO_2# must be converted to moles

#2.34g CO_2 x (1mol)/(44gCO_2) = 0.053 mols#

#P = 1atm#
#V= ??? L#
#n= 0.053 mols#
#R=0.0821 (atmL)/(molK)#
#T=273K#

#PV=nRT# becomes

#V = (nRT)/P#

#V = (0.053cancel(mols)(0.0821 (cancel(atm)L)/(cancel(mol)cancel(K)))(273cancelK))/(1atm)#

#V = 1.18 L#