#sf((a))#
#sf(2NaHCO_(3(s))rarrNa_2CO_(3(s))+CO_(2(g))+H_2O_((g)))#
#sf(K_p=p_(CO_2)xxp_(H_2O)=0.25color(white)(x)"Atm"^2)#
Since #sf(p_(CO_2)=p_(H_2O)# we can say:
#sf(p_(CO_2)^2=0.25)#
#:.##sf(p_(CO_2)=sqrt(0.25)=color(red)(0.50color(white)(x)"Atm"))#
and
#sf(p_(H_2O)=color(red)(0.50color(white)(x)"Atm"))#
#sf((b))#
We can get the initial moles #sf(n)# of #sf(NaHCO_3)# using:
#sf(n=m/M_r=13.6/84.0=0.162)#
Set up an ICE table based on moles:
#sf(" "2NaHCO_(3(g))rarrNa_2CO_(3(s))+CO_(2(g))+H_2O_((g)))#
#sf(I" "0.162" "0" "0" "0)#
#sf(C" "-2x" "+x" "+x" "+x)#
#sf(E" "(0.162-2x) " "x" "x" "x)#
We can find x from the volume of #sf(CO_2)#:
#sf(PV=n_(CO_2)RT)#
#:.##sf(n_(CO_2)=(PV)/(RT)=(0.50xx1.00)/(0.082xx398)=0.0153)#
#:.##sf(x=0.0153)#
#:.##sf(n_(Na_2CO_3)=0.0153)#
#:.##sf(m_(Na_2CO_3)=n_(Na_2CO_3)xxM_r=0.0153xx105.98=color(red)(1.621color(white)(x)g))#
#sf(n_(NaHCO_3)=0.162-2xx(0.0153)=0.1314)#
#:.##sf(m_(NaHCO_3)=n_(NaHCO_3)xxM_r=0.1314xx84.00=color(red)(11.04color(white)(x)g))#
#sf((c))#
#sf(2NaHCO_(3(g))rarrNa_2CO_(3(s))+CO_(2(g))+H_2O_((g)))#
#sf(2mol" "rarr" "1mol" "+1mol" "1mol)#
Converting to grams #sf(rArr)#
#sf(2xx84.00g" "rarr" "1mol" "mol)#
#sf(168g" "rarr" "1mol" "1mol)#
If all the #sf(NaHCO_3)# decomposes we can say that:
#sf(1g" "rarr" " 1/(168)mol" "1/(168)mol)#
#:.##sf(13.6g" "rarr" "13.6/(168)mol" "13.6/(168)mol)#
#sf(13.6g" "rarr" "0.081mol" "0.081mol)#
So the total moles of gas produced =#sf(2xx0.081=0.162)#
If the gases expand aqainst the atmosphere we can use the ideal gas expression:
#sf(PV=nRT)#
#:.##sf(V=(nRT)/(P)=(0.162xx0.082xx398)/(1.00)=color(red)(5.28color(white)(x)L))#
This ignores the volume taken up by any solids.
