Дано
n(HCL)=3 mol
Al
-------------------
V(H2)-?
3mol Xmol
2Al+6HCL-->2ALCL3+3H2
6mol 3mol
3/6=X/3
X=1.5 mol
V(H2)=n*Vm=1.5 *22.4=33.6 L
ответ 33.6 л
1)Ca+2H2O=Ca(OH)2+H2
2)Ca(OH)2+2HCl=CaCl2+2H2O
3)CaCl2+H2CO3=CaCO3+2HCl
4)CaCO3+2HNO3=Ca(NO3)2+H2O+CO2
5)Ca(NO3)2+2HNO2=Ca(NO2)2+2HNO3
<span>H3AsO4 + 3KOH → K3AsO4 + 3H2O
</span>3H+ +3OH-=3H2O
Ba(OH)2 + H2SO4(разб.) = BaSO4↓ + 2H2O
Ba2+ + 2OH− + 2H+ + SO42− = BaSO4 + 2H2O
<span>
2HNO3+Ba(OH)2=Ba(NO3)2+2H2O<span>
2H+ +2OH- =2H2O
</span></span>2HNO3(разб.) + Cu(OH)2 = Cu(NO3)2 + 2H2O
<span>2H++ Cu(OH)₂ = Cu2+ + 2H₂O</span>