Răspuns:
1-dizolvat
2- substantele dizolvate
3- pure
4-exces
5-constant
6-diferenta
1molN2......3molH2,,,,,,,,,,,,2molNH3
X.................3molH2......................y
X= 1molN2se consuma se raman neconsumati 1mol
y=2molNH3
A,Al=27g/mol
P=m,px100/m,i---> m,p= 101,25x80/100g=81gAl pur
niu=m/A= 81g/27g/mol=3molAl
din ecuatie, 2molAl.............3molH2
3mol....................x= 4,5molH2
miuCuO=80g/mol
m,CuCO3=124g/mol
niu,CuCO3= m/miu=12,4/124mol=0,1mol
din ecuatie, rezulta 0,1molCuO-teoretic
m,CuO,teoretic=niuxmiu=0,1x80g=8g
m,practic=8x90/100g=7,2g
c=mdx100/ms
md,acid=cxms/100=20xms/100
md,baza= cxms/100= 40x350/100g=140g
din ecuatie 36,5gHCl...........40gNaOH...........58,5gNaCL
X................................140g......................Y
x=127,75gHCl=md
127,75=20xms/100----> ms=638,75g
Y= 140X58,5/40g=204,75gNaCl
Explicație: