$n_{Zn}=52/65=0,8mol$

$a.Zn+2HCl\to ZnCl_2+H_2↑$

Theo pt :

$n_{H_2}=n_{Zm}=0,8mol$

$⇒V_{H_2}=0,8.22,4=17,92l$

b.Theo pt :

$n_{HCl}=2.n_{Zn}=2.0,8=1,6mol$

$⇒m_{HCl}=1,6.36,5=58,4g$

$⇒m_{ddHCl}=\dfrac{58,4}{10,95\%}≈533,33g$

c.Theo pt :

$n_{ZnCl_2}=n_{Zn}=0,8mol$

$⇒m_{ZnCl_2}=0,8.136=108,8g$

$m_{ddspu}=52+533,33-0,8.2=583,73g$

$⇒C\%_{ZnCl_2}=\dfrac{108,8}{583,73}=18,64\%$

$d.n_{Fe_2O_3}=80/160=0,5mol

                    $Fe_2O_3   +   H_2\overset{t^o}\to 2Fe+3H_2O$

Theo pt :       1 mol          3 mol

Theo đbài :   0,5 mol      0,8 mol

⇒Sau pư Fe2O3 dư

Theo pt :

$n_{fe}=2/3.n_{H_2}=2/3.0,8=\dfrac{8}{15}mol$

$⇒m_{Fe}=\dfrac{8}{15}.56=\dfrac{448}{15}g$

a,ta có phương trình:Zn+2HCl=>ZnCl2+H2

ta có mZn=52(g)=>nZn=$\frac{52}{65}$=0.8(mol) 

=>nH2=0.8(mol)=>vH2=0.8*22.4=17.92(lít)

b,nHCl=0.8*2=1.6(mol)=>mHCl=1.6*36.5=58.4(g)

=>mđ HCl=$\frac{58.4*100}{10.95}$≈533.33(g)

c,nZnCl2=0.8(mol)=>mZnCl2=0.8*(65+35.5*2)=108.8(g)

mdd ZnCl2=52+533.33-0.8*2=583.73(g)

=>C%=$\frac{108.8}{583.73}$*100=18.64% 

d,3H2+Fe2O3=>2Fe+3H2O

ta có mFe2O3=80(g)=>nFe2O3=$\frac{80}{56*2+16*3}$=0.5(mol) 

ta có tỉ lệ:nH2:nFe2O3=$\frac{0.8}{3}$<$\frac{0.5}{1}$(H2 hết,Fe2O3 dư)

=>nFe=$\frac{0.8*2}{3}$=0.533(mol) 

=>mFe=0.5333*56=29.87(g)