Получится борноэтиловый эфир.
1.1 2Al+6H2NO3=2Al(NO3)3 +3H2
1.2 Al2Cl3+3NaNO3=Al(NO3)3+3NaCl
1.3 2Al2O3+6H2NO3=2Al(HO3)3+3H2O
2.1 2Cu2O3+6HCl=2CuCl3+3H2O
2.2 Cu(OH)3+3NaCl=CuCl3+3NaOH
2.3 3CuPO4+3CaCl2=2CuCl3+Ca3(PO4)3
1)ZnCl2 + 2KOH = Zn(OH)2 + 2KCl
Mr(AIPO₄) =27+31 +16x4 =27+31+64=122
ω(AI)= 27÷122= 0.221 ω%(AI)=0.221×100% = 22.1%
ω(P) =31÷122 = 0.254 ω%(P)=0.254×100% = 25.4%
ω(O) =64÷122 = 0.525 ω%(O)=0.524×100% = 52.5%