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10,7g NH4Cl recationeaza cu hidroxidul de calciu
se cere
1)Ecuatia reactiei chimice
2)Calculati volumul de gaz masurat in conditii normale
3)Determinati concentratia solutiei finale


Răspuns :

1) 2 NH4Cl + Ca(OH)2 -> 2 NH3 + 2 H2O + CaCl2
2) In c.n un mol de gaz ocupa un volum de 22.4 L=dmcubi
M NH4Cl=14+4*1+35.5=53.5
nNH4Cl=m/M=10,7/53.5=0.2 moli

Pe rc. avem 0.2/2=x/2 moli NH3 => x=0.2 moli NH3

n=V/Vm. unde Vm=22.4 L=> V=nVm=0.2*22.4=44.8 L NH3

3) c=md/ms*100
ms=mH2O+mdCaCl2

nH2O de pe reactie=0.2 =>MH2O=0.2*18=9 g apa

y/1=0.2/2 raport molar din reactie, y-nr de moli de sarea CaCl2 => y=0.1 moli CaCl2 =>

mdCaCl2= n*M= 0,1* 111=11.1 g

ms=11.1+9=20.1
c=100*11.1/20.1=55, 22%