n_Zn=\(\dfrac{m}{M}=\dfrac{12,8}{65}=0,197\left(mol\right)\)

pthh:

Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂

0,197..0,394.....0,197...0,197(mol)

2H₂ + O₂ \(\rightarrow\) 2H₂O

0,197................0,197

m_H2O=n.M=0,197.18=3,546(g)

đề có sai chỗ nào ko vậy

mik thấy thay Zn bằng Cu hợp lí hơn

$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH : 

$n_{H_2} = n_{Zn} = \dfrac{12,8}{65} = 0,197(mol)$
$2H_2 + O_2 \xrightarrow{t^o} 2H_2O$
Theo PTHH : 

$n_{H_2O} = n_{H_2} = 0,197(mol)$
$m_{H_2O} = 0,197.18 = 3,546(gam)$

\(n_{Zn}=\dfrac{12,8}{65}=\dfrac{64}{325}\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ H_2+\dfrac{1}{2}O_2\underrightarrow{^{to}}H_2O\\ n_{H_2O}=n_{ZnCl_2}=n_{H_2}=n_{Zn}=\dfrac{64}{325}\left(mol\right)\\ m_{H_2O}=\dfrac{64}{325}.18\approx3,545\left(g\right)\\ m_{ZnCl_2}=\dfrac{64}{325}.136\approx26,782\left(g\right)\)

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2.......0.4.......................0.2\)

\(m_{Zn}=0.2\cdot65=13\left(g\right)\)

\(C\%_{HCl}=\dfrac{0.4\cdot36.5}{200}\cdot100\%=7.3\%\)

\(n_{CuO}=\dfrac{24}{80}=0.3\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(1..........1\)

\(0.3.........0.2\)

\(LTL:\dfrac{0.3}{1}>\dfrac{0.2}{1}\Rightarrow CuOdư\)

\(m_{CuO\left(dư\right)}=\left(0.3-0.2\right)\cdot64=6.4\left(g\right)\)

\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right);n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\\ a,Vì:\dfrac{0,3}{1}< \dfrac{0,8}{2}\Rightarrow HCldư\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\\ n_{HCl\left(dư\right)}=0,8-0,3.2=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{MgCl_2}=0,3.95=28,5\left(g\right)\\ m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)

a) \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

\(n_{HCl}=\dfrac{29.2}{36,5}=0,8\left(mol\right)\)

PTHH : 2Mg + 2HCl -> 2MgCl + H₂

Xét tỉ lệ \(\dfrac{0,3}{2}< \dfrac{0,8}{2}\)

=> HCl dư

=> \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

=> \(V_{MgCl}=0,15.22,4=3,36\left(l\right)\)

b) \(m_{H_2}=0,075.2=0,15\left(g\right)\\ m_{MgCl}=0,15.59,5=8,925\left(g\right)\)

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}=6,5+7,3-13,6=0,2\left(g\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl -->ZnCl₂ + H₂

____0,2<----------------------0,2

=> m_Zn = 0,2.65 = 13 (g)

m_Cu = m_{rắn không tan} = 19,5 (g)

\(\left\{{}\begin{matrix}\%Zn=\dfrac{13}{13+19,5}.100\%=40\%\\\%Cu=\dfrac{19,5}{13+19,5}.100\%=60\%\end{matrix}\right.\)

`n_(H_2)=4,48/22,4=0,2 (mol)`

Ta có PTHH: `Zn+2HCl --> ZnCl_2 +H_2`

Theo PT:       `1`--------------------------------`1`

Theo đề:       `0,2`------------------------------`0,2`

`m_(Zn)=0,2.65=13(g)`

Vì `Cu` không phản ứng với `HCl` nên `m_(chất rắn không tan)=m_(Cu)=19,5(gam)`

`%Zn=13/(13+19,5) .100%=40%`

`%Cu=100%-40%=60%`

a)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,2-->0,4------>0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl(PTHH) = 0,4.36,5 = 14,6 (g)

=> \(m_{HCl\left(tt\right)}=\dfrac{14,6.120}{100}=17,52\left(g\right)\)

c) 

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) => H₂ hết, O₂ dư

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

           0,2--->0,1------->0,2

=> m_H2O = 0,2.18 = 3,6 (g)

m_O2(dư) = (0,2 - 0,1).32 = 3,2(g)

nZn = 13/65 = 0,2 (mol)

PTHH: Zn + 2HCl -> ZnCl2 + H2

Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,2

VH2 = 0,2 . 22,4 = 4,48 (l)

mHCl = (0,4 . 36,5)/(100% + 20%) = 73/6 (g)

nO2 = 4,48/22,4 = 0,2 (mol)

PTHH: 2H2 + O2 -> (t°) 2H2O

LTL: 0,2/2 < 0,2 => O2 dư

nH2O = nH2 = 0,2 (mol)

mH2O = 0,2 . 18 = 3,6 (g)

V = 0,1 : 0,5 = 0,2 (l)

_​                 a                       ->                 3/2a

Fe  + H₂SO₄  -> FeSO₄  + H₂

a                        ->             a

c

\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,15(mol)\\ \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{HCl} = \dfrac{8,1}{36,5} = \dfrac{81}{365}(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ \dfrac{n_{Al}}{2} = 0,05 > \dfrac{n_{HCl}}{6} = \dfrac{27}{730} \to Al\ dư\\ n_{Al\ pư} = \dfrac{1}{3}n_{HCl} = \dfrac{27}{365}(mol)\\ \)

\(m_{Al\ dư} = 2,7 - \dfrac{27}{265}.27 = 0,703(gam)\)