\(n_{Cl_2}=a\left(mol\right),n_{O_2}=b\left(mol\right)\)

\(n_{hh}=a+b=0.25\left(mol\right)\left(1\right)\)

BTKL : 

\(m_{khí}=23-7.2=15.8\left(g\right)\)

\(\Rightarrow71a+32b=15.8\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.05\)

\(2M+nCl_2\underrightarrow{^{^{t^0}}}2MCl_n\)

\(4M+nO_2\underrightarrow{^{^{t^0}}}2M_2O_n\)

\(n_M=\dfrac{0.4}{n}+\dfrac{0.2}{n}=\dfrac{0.6}{n}\left(mol\right)\)

\(M_M=\dfrac{7.2}{\dfrac{0.6}{n}}=12n\)

\(n=2\Rightarrow M=24\)

\(M:Mg\)

Gọi $n_{Cl_2} = a ; n_{O_2} = b \Rightarrow a + b = 0,25(1)$
Bảo toàn khối lượng :

$7,2 + 71a + 32b = 23(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,05

Gọi n là hóa trị M

$2M + nCl_2 \to 2MCl_n$
$4M + nO_2 \xrightarrow{t^o} 2M_2O_n$

Theo PTHH : 

$n_M = \dfrac{2}{n}n_{Cl_2} + \dfrac{4}{n}n_{O_2} = \dfrac{0,6}{n}$

$\Rightarrow \dfrac{0,6}{n}.M = 7,2$
$\Rightarrow M = 12n$

Với n = 2 thì $M = 24(Magie)$

a)Thí nghiệm 2 :

$2R + O_2 \xrightarrow{t^o} 2RO$

n O2 = 1,12/22,4 = 0,05(mol)

n R < 2 n O2 = 0,1(mol)

=> R > 3/0,1 = 30(1)

Thí nghiệm 2 : n H2 = 5,6/22,4 = 0,25(mol)

n R = a(mol)

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$R + H_2SO_4 \to RSO_4 + H_2$

n Fe = 0,25 - a(mol)

Suy ra :

(0,25 - a).56 + Ra = 11,6

=> R = (56a - 2,4)/a

Mà 0 < a < 0,25

Suy ra : R < 46,6(2)

Từ (1)(2) suy ra : 30 < R < 46,6

Suy ra R là Ca(R = 40)

b) (0,25 -a)56 + 40a = 11,6

=> a = 0,15

%m Ca = 0,15.40/11,6   .100% = 51,72%
%m Fe = 100% -51,72% = 48,28%

\(a,m_C=48\left(g\right)\rightarrow n_C=\dfrac{m_C}{M_C}=\dfrac{48}{12}=4\left(mol\right)\)

\(V_{O_2}=44,8\left(l\right)\rightarrow n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)

\(PTHH:C+O_2\underrightarrow{t^o}CO_2\)

\(pt:\)      \(1mol\)  \(1mol\)

\(đb:\)      \(4mol\)  \(2mol\)

Xét tỉ lệ:

\(\dfrac{n_{C\left(đb\right)}}{n_{C\left(pt\right)}}=\dfrac{4}{1}=4>\dfrac{n_{O_2\left(đb\right)}}{n_{O_2\left(pt\right)}}=\dfrac{2}{1}=2\)

\(\Rightarrow\) \(O_2\) hết, \(C\) dư.

\(b,PTHH:C+O_2\underrightarrow{t^o}CO_2\)

\(pt:\)                  \(1mol\)   \(1mol\)

\(đb:\)                  \(2mol\)   \(2mol\)

\(\Rightarrow m_{CO_2}=n_{CO_2}.M_{CO_2}=2.\left(1.C+2.O\right)=2.\left(1.12+2.16\right)=88\left(g\right)\)

\(a.n_C=\dfrac{48}{12}=4\left(mol\right);n_{O_2}=\dfrac{44,8}{22,4}=2\left(mol\right)\\ C+O_2\xrightarrow[t^0]{}CO_2\)

Theo pt:\(\dfrac{4}{1}>\dfrac{2}{1}\Rightarrow C\) dư, O₂ pư hết

\(b.C+O_2\xrightarrow[t^0]{}CO_2\\ \Rightarrow n_{CO_2}=n_{O_2}=2mol\\ m_{CO_2}=2.44=88\left(g\right)\)

\(n_{KMnO_4}=\dfrac{7,9}{158}=0,05\left(mol\right)\)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

            0,05-------------------------------->0,025

=> n_{O2(cần dùng)} = \(\dfrac{0,025.80}{100}=0,02\left(mol\right)\)

PTHH: 4R + nO₂ --t^o--> 2R₂O_n

       \(\dfrac{0,08}{n}\)<-0,02

=> \(M_R=\dfrac{0,96}{\dfrac{0,08}{n}}=12n\left(g/mol\right)\)

- Nếu n = 1 => M_R = 12 (Loại)

- Nếu n = 2 => M_R = 24 (Mg)

- Nếu n = 3 => M_R = 36 (Loại)

- Nếu n = \(\dfrac{8}{3}\) => M_R = 32 (Loại)

Vậy R là Mg

\(2KMnO_4 \rightarrow K_2MnO_4+MnO_2+O_2 \uparrow\\ n_{KMnO_4}=\dfrac{7,9}{158}=0,05(mol)\\ \Rightarrow n_{O_2} = \dfrac{0,05}2=0,025(mol)\\ n_{{O_2}_{\text{Cần dùng}}}=0,025.80=0,02(mol)\\4R+nO_2 \rightarrow 2R_2O_n\\ \Rightarrow n_R=\dfrac{0,02.4}{n}=\dfrac{0,08}n (mol)\\ \Rightarrow M_R=\dfrac{0,96:0,08}n=12n\\ \text{ Kẻ bảng biện luận }\\ \begin{array}{|c|c|c|}\hline \text{n=1}&\text{n=2}&\text{n=3}\\\hline \text{M=12(loại)}&\text{M=24(nhận)(Mg)}&\text{36(loại)}\\\hline\end{array}\\\text{Vậy M là Mg} \)

Ta có hệ: \(\left\{{}\begin{matrix}m_R+m_{Mg}=6,3\\\dfrac{m_R}{m_{Mg}}=\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_R=2,7g\\m_{Mg}=3,6g\end{matrix}\right.\)

\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)

\(2Mg+O_2\underrightarrow{t^o}2MgO\)

0,15      0,075

Mà \(\Sigma n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(\Rightarrow n_{O_2\left(R\right)}=0,15-0,075=0,075mol\)

\(4R+nO_2\underrightarrow{t^o}2R_2O_n\)

\(\dfrac{2,7}{R}\)   0,075

\(\Rightarrow\dfrac{2,7}{R}\cdot n=4\cdot0,075\Rightarrow9n=A\)

Nhận thấy n=3 thỏa mãn\(\Rightarrow R=27\Rightarrow Al\)

Câu b khuất đề nên mình ko làm đc nhé!!!

a: \(2Mg+O_2\rightarrow2MgO\)

\(m_{O_2}=15-9=6\left(g\right)\)

b: \(C+O_2\rightarrow CO_2\)

m=22-16=6(g)