\(n_{H_2}=\dfrac{33.6}{22.4}=1.5\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(1.5........1.5.....................1.5\)

\(m_{Fe}=1.5\cdot56=84\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{1.5}{0.5}=3\left(M\right)\)

\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\)

\(1.5........0.75\)

\(V_{O_2}=0.75\cdot22.4=16.8\left(l\right)\)

Giúp e câu c vs ạ!

\(n_{Al}=\dfrac{8.1}{27}=0.3\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.3..........0.9...........0.3..........0.45\)

\(C_{M_{HCl}}=\dfrac{0.9}{0.1}=9\left(M\right)\)

\(m_{AlCl_3}=0.3\cdot133.5=40.05\left(g\right)\)

\(V_{H_2}=0.45\cdot22.4=10.08\left(l\right)\)

a: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

c: \(n_{Al}=\dfrac{8.1}{27}=0.3\left(mol\right)\)

=>nAlCl3=0,3(mol)

\(m=0.3\cdot\left(27+35.5\cdot3\right)=40.05\left(g\right)\)

d: nHCl=0,9(mol)

=>nH2=0,45mol

\(V=0.45\cdot22.4=10.08\left(lít\right)\)

\(n_{Al}=\dfrac{8.1}{27}=0.3\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.3..........0.9...........0.3..........0.45\)

\(C_{M_{HCl}}=\dfrac{0.9}{0.1}=9\left(M\right)\)

\(m_{AlCl_3}=0.3\cdot133.5=40.05\left(g\right)\)

\(V_{H_2}=0.45\cdot22.4=10.08\left(l\right)\)

\(n_{Zn}=\dfrac{19.5}{65}=0.3\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

\(m_{ZnSO_4}=0.3\cdot161=48.3\left(g\right)\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

$n_{FeCl_2} = \dfrac{25,4}{127} = 0,2(mol)$

Gọi $n_{FeCl_3} = a(mol)$

Bảo toàn nguyên tố với Cl  :

$n_{HCl} = 2n_{FeCl_2} + 3n_{FeCl_3} = 0,4 + 3a(mol)$

$n_{H_2O} = \dfrac{1}{2}n_{HCl} = 0,2 + 1,5a(mol)$

Bảo toàn khối lượng : 

$m_{hh} + m_{HCl} = m_{FeCl_2} + m_{FeCl_3} + m_{H_2O}$

$\Rightarrow 30,4 + (0,4 + 3a).36,5 = 25,4 + 162,5a + (0,2 + 1,5a).18$
$\Rightarrow a = 0,2$

$m = 0,2.162,5 = 32,5(gam)$

\(n_{FeCl_2}=\dfrac{25.4}{127}=0.2\left(mol\right)\)

\(n_{HCl}=a\left(mol\right)\)

Bảo toàn Cl : 

\(n_{HCl}=2n_{FeCl_2}+3n_{FeCl_3}\)

\(\Rightarrow n_{FeCl_3}=\dfrac{a-2\cdot0.2}{3}=\dfrac{a-0.4}{3}\left(mol\right)\)

Bảo toàn H : 

\(n_{H_2O}=\dfrac{n_{HCl}}{2}=\dfrac{a}{2}\left(mol\right)\)

BTKL : 

\(m_{hh}+m_{HCl}=m_{FeCl_2}+m_{FeCl_3}+m_{H_2O}\)

\(\Rightarrow30.4+36.5a=25.4+\dfrac{a-0.4}{3}\cdot162.5+\dfrac{a}{2}\cdot18\)

\(\Rightarrow a=1\)

\(m_{FeCl_3}=\dfrac{1-0.4}{3}\cdot162.5=32.5\left(g\right)\)

`m_{hh} = 0,2.32 + 0,35.46 = 22,5 (g)`

$n_{H_2} = \dfrac{0,3072}{2} = 0,1536(mol)$
Gọi $n$ là hoá trị của kim loại $M$

$2M + 2nHCl \to 2MCl_n + nH_2$
Theo PTHH : 

$n_M = \dfrac{2}{n}.n_{H_2} = \dfrac{0,3072}{n}(mol)$
$\Rightarrow \dfrac{0,3072}{n}.M = 6,144$
$\Rightarrow M = 20n$

Với n = 2 thì M = 40

Vậy M là kim loại Canxi

a) $Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$

b) $n_{Al_2O_3} = \dfrac{10,2}{102} = 0,1(mol)$

Theo PTHH : $n_{Al_2(SO_4)_3} = n_{Al_2O_3} = 0,1(mol)$
$m_{Al_2(SO_4)_3} = 0,1.342 = 34,2(gam)$

c) $n_{H_2SO_4} =3n_{Al_2O_3} = 0,3(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,3}{1} = 0,3(lít)$

Ta có: \(n_{CO_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

PT: \(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

Theo PT: \(n_{KOH}=2n_{CO_2}=0,05\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

\(n_{CO_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

\(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

0,025 → 0,05

Đổi: \(100ml=0,1l\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)