Giải thích các bước giải:

1/.

a/.

* $CTHH$: $N_2$

$PTK_{N_2}=14.2=28(đvC)$

b/.

* $CTHH$: $Fe_2O_3$

$PTK_{Fe_2O_3}=56.2+16.3=160(đvC)$

c/.

* $CTHH$: $Al_2(SO_4)_3$

$PTK_{Al_2(SO_4)_3}=27.2+(32.1+16.4).3=342(đvC)$

2/.

a/. $4P+5O_2$ $\xrightarrow{t^o}$ $2P_2O_5$

Tỉ lệ: $4:5:2$

c/. $MgO+2HCl→MgCl_2+H_2O$

Tỉ lệ: $1:2:1:1$

b/. $Ca+Cl_2$ $\xrightarrow{t^o}$ $CaCl_2$

Tỉ lệ: $1:1:1$

d/. $2Al+Fe_2O_3$ $\xrightarrow{t^o}$ $Al_2O_3+2Fe$

Tỉ lệ: $2:1:1:2$

3/.

a/.

PTHH:

$Zn+HCl→ZnCl_2+H_2↑$

b/. 

Công thức khối lượng:

$m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}↑$

c/.

$m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}↑$

$6,5+7,3=13,6+m_{H_2}$

⇒ $m_{H_2}=6,5+7,3-13,6=0,2g$

d/.

$n_{H_2}=$ `(0,2)/2` $=0,1mol$

$V_{H_2(đktc)}=0,1.22,4=2,24 lít$