#50g# of pure #CaCO_3# is heated, liberated #CO_2# reacted 0.4 mol of moist ammonia to yield only #(NH_4_2CO_3#. Find the volume of 3 #CO_2# left after the reaction at STP?

A) zero
B) #6.72L#
C) #2.24L#
D)indeterminate from this data

1 Answer
Dec 4, 2017

6.72L

Explanation:

Chemical reaction: 2# NH_3 + + CO_2 rArr NH_2COONH_4#

NH3 reacts with CO2 to give Ammonium carbamate.

Moles of #CO_2# taken:50/100=0.5 mol

Moles of #NH_3# reacted#rarr#0.4 mol{given}

According to the reaction:
2 moles of # NH_3# react with#rarr#1 mole of# CO_2#

For the reaction of 0.4 mole of #NH_3 rarr#1/2#xx#0.4 mol of CO2 reacted, #rarr#0.4#xx#0.5 moles of #CO_2#

moles of #CO_2# left after reaction= moles taken# -#moles reacted
# =#0.5#-#0.4#xx#0.5#=#0.5(1#-#0.4)#rArr#0.6/2 #rArr#0.3

moles of #CO_2# left after reaction #rarr#0.3 moles

Volume of 0.3 moles of #CO_2=#22.4L#xx#0.3{1 mole#rArr#22.4L}
#rArr#6.72L