Bài 1:

1)2Na + 2 H2O → 2NaOH +H2 . NaOH:Natri hiđroxit

2)CaO + H2O → Ca(OH)2 . Ca(OH)2:Canxi hiđroxit

3)P2O5 +3 H2O→ 2 H3PO4:Axit Photphoric

4)Zn + H2SO4 → ZnSO4 +H2 .ZnSO4:Kẽm sunfat

5)Fe + 2HCl → FeCl2 +H2 .FeCl2:sắt(II) clorua 

6)2Al + 3 H2SO4 → Al2(SO4)3 +3 H2 .Al2(SO4)3:nhôm sunfat

Bài 2:

a)2Al + 6HCl→2AlCl3 +3H2

$n_{Al}$=$\frac{10,8}{27}$ =0,4(mol)

theo pt:2molAl→3molH2

theo đề:0,4molAl→0,6molH2

$V_{H2}$=0,6*22,4=13,44(l)

b)theo pt:2mol Al→6molHCl

theo đề:0,4molAl→1,2molHCl

=>$C_{M_{HCl}}$=$\frac{1,2}{0,5}$ =2,4(M)

Đáp án:

$\rm Bài \ 1 \ : \\ 1) \ 2Na + 2H_2O \to 2NaOH + H_2 \ \ | \ NaOH \ : \ Bazơ \ : \ Natri \ hiđroxit \\ 2) \ CaO + H_2O \to Ca(OH)_2 \ \ -Bazơ \ : \ Canxi \ hiđroxit \\ 3) \ P_2O_5 + 3H_2O \to 2H_3PO_4 \ - Axit \ : \ Axit \ photphoric \\ 4) \ Zn + H_2SO_4 \to ZnSO_4 + H_2 \ -Muối \ : \ Kẽm \ sunfat \\ 5) \ Fe + 2HCl \to FeCl_2 + H_2 \ -Muối \ : \ Sắt \ (II) \ clorua \\ 6) \ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2 \ - \ Muối \ : \ Nhôm \ sunfat \\ Bài \ 2 \ : \\ a) \ 2Al + 6HCl \to 2AlCl_3 + 3H_2 \\ Ta \ có \ : \ n_{Al}=\dfrac{10,8}{27}=0,4 \ (mol) \\ \to n_{H_2}=\dfrac{0,4.3}{2}=0,6 \ (mol) \\ \to V_{H_2}=0,6.22,4=13,44 \ (l) \\ b) \\ 500ml  \ = \ 0,5l \\ Ta \ có \ : \ n_{HCl}=\dfrac{0,4.6}{2}=1,2 \ (mol) \\ \to C_{M_{HCl}}=\dfrac{1,2}{0,5}=2,4 \ (M)$