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.
