\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)

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

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

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{Cu}=0.2\cdot64=12.8\left(g\right)\)

\(2Cu+O_2\underrightarrow{^{t^0}}2CuO\)

\(0.2......0.1\)

\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)

\(n_{FeO}=\dfrac{32}{72}=\dfrac{4}{9}\left(mol\right)\)

\(FeO+H_2\underrightarrow{^{t^0}}Fe+H_2O\)

\(\dfrac{4}{9}.....\dfrac{4}{9}....\dfrac{4}{9}\)

\(V_{H_2}=\dfrac{4}{9}\cdot22.4=10\left(l\right)\)

\(m_{Fe}=\dfrac{4}{9}\cdot56=24.89\left(g\right)\)

\(a) n_{FeO} = \dfrac{32}{72} = \dfrac{4}{9}(mol)\\ FeO + H_2 \xrightarrow{t^o} Fe + H_2O\\ n_{Fe} = n_{H_2} = n_{FeO} = \dfrac{4}{9}(mol)\\ V_{H_2} = \dfrac{4}{9}.22,4 = 9,95(lít)\\ b) m_{Fe}= \dfrac{4}{9}.56 = 24,8(gam)\)

\(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
pthh: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
        0,1                 0,1       0,1
=>\(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{H_2O}=0,1.18=1,8\left(g\right)\end{matrix}\right.\)

a)

$n_{Mg} = \dfrac{7,2}{24} = 0,3(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
Theo PTHH : $n_{H_2} = n_{Mg} = 0,3(mol)$
$V_{H_2} = 0,3.24,79 = 7,437(lít)$

b)

$n_{MgCl_2} = n_{Mg} = 0,3(mol)$
$m_{MgCl_2} = 0,3.95 = 28,5(gam)$

c)

Gọi CTHH oxit là $R_2O_n$
$R_2O_n + nH_2 \xrightarrow{t^o} 2R + nH_2O$

Theo PTHH : $n_{oxit} = \dfrac{1}{n}.n_{H_2} = \dfrac{0,3}{n}(mol)$
$\Rightarrow \dfrac{0,3}{n}(2R + 16n) = 21,6$
$\Rightarrow R = 28n$

Với n = 2 thì R = 56(Fe)

Vậy oxit là $FeO$

nMg=7,224=0,3(mol)nMg=7,224=0,3(mol)
Mg+2HCl→MgCl2+H2Mg+2HCl→MgCl2+H2
nH2=nMg=0,3(mol)nH2=nMg=0,3(mol)
VH2=0,3.24,79=7,437(lít)VH2=0,3.24,79=7,437(lít)

b)

$n_{MgC{l}_2}=n_{Mg}=0,3\left(mol\right)$

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

\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)

tỉ lệ       :1          2           1              1

số mol  :0,3       0,6        0,3           0,3

\(V_{H_2}=0,3.24,79=7,437\left(l\right)\)

b)\(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)

c) Gọi CTHH của oxide là: \(R_xO_y\)

\(PTHH:R_xO_y+xH_2\xrightarrow[]{}xR+xH_2O\)

Theo phương trình ta có:\(2,48.\dfrac{1}{2}.\dfrac{R}{x}=21,6\)

\(\Rightarrow R_xO_y=80\left(đvc\right)\)

\(\Rightarrow R=56\Rightarrow Fe\)

Với x=2 và y=2 

thì CTHH có dạng \(FeO\)

nHgO = 21,7 : 217 = 0,1 (mol) 
pthh : HgO + H2 -t--> H2O + Hg
         0,1---->0,1--------------->0,1 (MOL) 
=> mHg = 0,1 . 201 = 20,1 (G) 
=> VH2 = 0,1 . 22,4 = 2,24 (L) 

a)

$n_{Mg} = \dfrac{7,2}{24} = 0,3(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
Theo PTHH : $n_{H_2} = n_{Mg} = 0,3(mol)$
$V_{H_2} = 0,3.24,79 = 7,437(lít)$

b)

$n_{MgCl_2} = n_{Mg} = 0,3(mol)$
$m_{MgCl_2} = 0,3.95 = 28,5(gam)$

c)

Bản chất : $H_2 + O_{oxit} \to H_2O$

$n_{O\ trong\ oxit} = n_{H_2} = 0,3(mol)$
$\Rightarrow m_{Fe} = m_{oxit} - m_O = 17,4 - 0,3.16 = 12,6(gam)$
$\Rightarrow n_{Fe} = 0,225(mol)$

Ta có : 

$n_{Fe} : n_O = 0,225 : 0,3 = 3 : 4$

Vậy CTHH của oxit là $Fe_3O_4$

a, \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)

b, \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

\(n_{Fe}=n_{FeO}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

c, \(n_{H_2}=n_{FeO}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(n_{Ag_2O}=\dfrac{23.2}{232}=0.1\left(mol\right)\)

\(Ag_2O+H_2\underrightarrow{^{t^0}}2Ag+H_2O\)

\(0.1......0.1.........0.2\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{Ag}=0.2\cdot108=21.6\left(g\right)\)

\(4Ag+O_2\underrightarrow{^{t^0}}2Ag_2O\)

\(0.2.....0.05\)

\(V_{kk}=5V_{O_2}=5\cdot0.05\cdot22.4=5.6\left(l\right)\)