What is the the volume occupied by 2.34 g of carbon dioxide gas at STP?

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Answer 1 · 2016-05-06T20:32:55

#V = 1.18 L#

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#

Answer 2 · 2016-05-07T02:17:48

#1.18 L#

Alternative solution:

Calculate how many moles #2.34g# #CO_2# is

#2.34g * (1mol)/(44g CO_2) = 0.053 mol#

Simple mole ratios tell us that #1# mol of any gas at STP is #22.4L#, but since we only have #0.053 mol#, we times #22.4L/(mol)# by #0.053mol.#

#(22.4L)/cancel(mol) * 0.053 cancel(mol) = 1.18L#