How many liters of carbon dioxide can be produced if 37.8 grams of carbon disulfide react with excess oxygen gas at 28.85 degrees Celsius and 1.02 atmospheres?
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Carbon disulfide reacts with oxygen to form carbon dioxide and sulfur dioxide.
#CS_2 + O_2 -> CO_2 + SO_2#
Start by finding the number of moles of the carbon dioxide produced.
#37.8 \ "g" \ CS_2 xx ( 1\ "mol."\ CS_2)/(76.1407 \ "g" \ CS_2)xx ("1 mol."\ CO_2)/( 1 \ "mol."\ CS_2)#
#37.8 \cancel ( "g" \ CS_2) xx ( color(red) cancel( 1\ "mol."\ CS_2))/(76.1407 \cancel( "g" \ CS_2))xx ("1 mol."\ CO_2)/(color (red) cancel( 1\ "mol."\ CS_2))#
#0.496 \ mol. \ CO_2#
Now use the ideal gas equation to figure out the the volume of the #CO_2# produced.
# P*V= n*R*T#
Where
#P" "# is the pressure expressed in atm.
#V" "# is the volume expressed in L.
#n" " # is the number of moles.
#R" "# is the universal gas constant, it has a value of #0.0821\ L* atm. mol^-1*K^-1#
#T" " #is the Kelvin temperature.
Now rearrange the ideal gas equation and solve for V.
# V= (n*R*T)/P#
# V= (0.496\ molxx0.0821\ L*atm*mol^-1*K^-1 xx 302.00 \ K)/ (1.02\ atm)#
#V= (0.496\cancel( mol)xx0.0821\ L*cancel(atm)*cancel(mol^-1)*cancel(K^-1)xx 302.00 \cancel(K))/ (1.02\ cancel(atm))#
#V= 12.1 \ L#
