$FeO + CO \xrightarrow{t^o} Fe + CO_2$

Theo PTHH :  $n_{FeO} = n_{CO\ pư} = n_{Fe} = n_{CO_2} = a(mol)$

$\Rightarrow m_{giảm} = m_{FeO} - m_{Fe} = 72a -56a = 16a = 1,6(gam)$
$\Rightarrow a = 0,1(mol)$

$m_{Fe} = 0,1.56 = 5,6(gam)$
$n_{CO\ dư} = 0,2 - 0,1 = 0,1(mol)$
$\%V_{CO\ dư} = \%V_{CO_2} = \dfrac{0,1}{0,1 + 0,1}.100\% = 50\%$

Cu+Cl₂->CuCl₂

Bài này cũng dễ mà sao bạn cho số không đẹp tí nào zậy, rắc rối

\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Ba + 2H₂O ---> Ba(OH)₂ + H₂

            0,3<-------------0,3<---------0,3

=> m_Ba = 0,3.137 = 41,1 (g)

=> m_K2O = 59,9 - 41,1 = 18,8 (g)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{41,1}{59,9}.100\%=68,61\%\\\%m_{K_2O}=100\%-68,61\%=31,39\%\end{matrix}\right.\)

\(b,n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)

PTHH: K₂O + H₂O ---> 2KOH

          0,2----------------->0,4

Các chất tan trong dd sau phản ứng: KOH, Ba(OH)₂

\(\rightarrow\left\{{}\begin{matrix}m_{KOH}=0,4.56=22,4\left(g\right)\\m_{Ba\left(OH\right)_2}=0,3.171=51,3\left(g\right)\end{matrix}\right.\)

\(n_{Cu} = a ; n_{Al} = b ; n_{Fe} = c(mol)\\ \Rightarrow 64a + 27b + 56c = 28,6(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\\ \text{Mặt khác} : n_{O_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 4Fe + 3O_2 \xrightarrow{t^o} 2Fe_2O_3\\ \)

Ta có :

\(\dfrac{n_X}{n_{O_2}}=\dfrac{a+b+c}{0,5a +0,75b + 0,75c} = \dfrac{0,6}{0,4}(3)\\ (1)(2)(3)\Rightarrow a = \dfrac{317}{1460} ; b = \dfrac{121}{365}; c = \dfrac{15}{146}\\ \%m_{Cu} = \dfrac{\dfrac{317}{1460}.64}{28,6}.100\% = 48,59\%\\ \%m_{Al} = \dfrac{\dfrac{121}{365}.27}{28,6}.100\% = 31,3\%\\ \%m_{Fe} = 100\% - 41,59\% - 31,3\% = 27,11\%\)

\(n_{Fe} =\dfrac{11,2}{56} = 0,2(mol)\\ \)

Fe + 2HCl → FeCl₂ + H₂

0,2.....0,4.........0,2........0,2..............(mol)

Vậy :

V = 0,2.22,4 = 4,48(lít)

\(m_{FeCl_2} = 0,2.127=25,4(gam)\)

\(m_{HCl} = 0,4.36,5 = 14,6(gam)\)

PTHH: Fe+2HCl → FeCl2+H2

a, nFe=m:M=11,2:56=0,2 mol

Theo PTHH, nFe=nH2=0,2 mol

VH2=n.22,4=0,2.22,4=4,48 lít

b, Theo PTHH, nFeCl2=nFe=0,2

mFeCl2=n.M=0,2.127=25,4 g 

c,

Theo PTHH, nHCl=2nFe=0,4 mol

mHCl=n.M=0,4.36,5=14,6 g

n_Fe=11,2/56=0,2(mol)

Fe + 2HCl ----->FeCl₂ + H₂

TPT:n_HCl=2.n_Fe=2.0,2=0,4(mol)

        m_HCl=0,4.36,5=14,6(g)

TPT:n_FeCl2=n_Fe=0,2(mol)

       m_FeCl2=0,2.127=25,4(g)

TPT:n_H2=n_Fe=0,2(mol)

       v_H2=0,2.22,4=4,48(lít)