Vì M có hóa trị IV `=>` CTHH của oxit là `MO_2`

$\rm \Rightarrow \%M = \dfrac{M_M}{M_M + 32} .100\% = 50\%$
$\rm \Rightarrow M_M = 32 (g/mol)$
Vậy M là nguyên tố lưu huỳnh (S)

a) \(n_{C_2H_4}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PTHH: \(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)

              0,2---->0,6---->0,4---->0,4

\(\Rightarrow V_{O_2}=0,6.24,79=14,874\left(l\right)\)

b) \(V_{CO_2}=0,4.22,4=9,916\left(l\right)\)

c) \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                       0,4------>0,4

\(\Rightarrow\left\{{}\begin{matrix}x=m_{CaCO_3}=0,4.100=40\left(g\right)\\y=m_{b\text{ình}.t\text{ăng}}=m_{CO_2}+m_{H_2O}=0,4.44+0,4.18=24,8\left(g\right)\end{matrix}\right.\)

a) \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right);n_{HCl}=\dfrac{36,5.10\%}{36,5}=0,1\left(mol\right)\)

PTHH:           \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ban đầu:       0,15      0,1

Pư:               0,05<----0,1

Sau pư:         0,1         0          0,05       0,05

\(\Rightarrow V=V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b) \(m_{dd.sau.p\text{ư}}=36,5+0,05.24-0,05.2=37,6\left(g\right)\)

dd sau phản ứng có chứa chất tan duy nhất là MgCl₂ 

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%=12,633\%\)

a) 

\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\) (1)

\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\) (2)

\(n_{KCl}=\dfrac{0,894}{74,5}=0,012\left(mol\right);m_B=\dfrac{0,894}{8,132\%}=11\left(g\right)\)

Gọi \(n_{O_2\left(sinh.ra\right)}=a\left(mol\right)\Rightarrow n_{kk}=3a\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2}=3a.80\%=2,4a\left(mol\right)\\n_{O_2}=a+\left(3a-2,4a\right)=1,6a\left(mol\right)\end{matrix}\right.\)

\(n_C=\dfrac{0,528}{12}=0,044\left(mol\right)\)

\(C+O_2\xrightarrow[]{t^o}CO_2\) (3)

Vì hỗn hợp D gồm 3 khí và O₂ chiếm 17,083%

\(\Rightarrow D:CO_2,O_{2\left(d\text{ư}\right)},N_2\)

BTNT C: \(n_{CO_2}=n_C=0,044\left(mol\right)\)

BTNT O: \(n_{O_2\left(d\text{ư}\right)}=n_{O_2\left(b\text{đ}\right)}-n_{CO_2}=1,6a-0,044\left(mol\right)\)

\(\Rightarrow\%V_{O_2}=\%n_{O_2}=\dfrac{1,6a-0,044}{1,6a-0,044+0,044+2,4a}.100\%=17,083\%\)

\(\Leftrightarrow a=0,048\left(mol\right)\left(TM\right)\)

ĐLBTKL: \(m_A=m_B+m_{O_2}=11+0,048.32=12,536\left(g\right)\)

Theo PT (2): \(n_{KClO_3}=n_{KCl}=0,012\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{KClO_3}=\dfrac{0,012.122,5}{12,536}.100\%=11,63\%\\\%m_{KMnO_4}=100\%-11,63\%=88,37\%\end{matrix}\right.\)

b) Theo PT (2): \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4\left(p\text{ư}\right)}+\dfrac{3}{2}n_{KClO_3}\)

\(\Rightarrow n_{KMnO_4\left(p\text{ư}\right)}=2.\left(0,048-\dfrac{3}{2}.0,012\right)=0,06\left(mol\right)\)

\(n_{KMnO_4\left(b\text{đ}\right)}=\dfrac{12,536-0,012.122,5}{158}=0,07\left(mol\right)\)

\(\Rightarrow n_{KMnO_4\left(d\text{ư}\right)}=0,07-0,06=0,01\left(mol\right)\)

\(n_{KCl}=\dfrac{74,5}{74,5}+0,012=1,012\left(mol\right)\)

Theo PT (1): \(n_{K_2MnO_4}=n_{MnO_2}=\dfrac{1}{2}.n_{KMnO_4\left(p\text{ư}\right)}=0,03\left(mol\right)\)

PTHH: 

\(2KMnO_4+10KCl+8H_2SO_4\rightarrow6K_2SO_4+2MnSO_4+5Cl_2+8H_2O\) (4)

\(K_2MnO_4+4KCl+4H_2SO_4\rightarrow3K_2SO_4+MnSO_4+2Cl_2+4H_2O\) (5)

\(MnO_2+2KCl+2H_2SO_4\rightarrow MnSO_4+K_2SO_4+Cl_2+2H_2O\) (6)

\(2KCl+H_2SO_4\xrightarrow[]{t^o}K_2SO_4+2HCl\) (7)

Theo PT (4), (5), (6): \(n_{KCl\left(p\text{ư}\right)}=5n_{KMnO_4\left(d\text{ư}\right)}+4n_{K_2MnO_4}+2n_{MnO_2}=0,23\left(mol\right)< 1,012\left(mol\right)=n_{KCl\left(b\text{đ}\right)}\)

`=> KCl` dư

Theo PT (4), (5), (6): \(n_{Cl_2}=\dfrac{1}{2}.n_{KCl\left(p\text{ư}\right)}=0,115\left(mol\right)\)

\(\Rightarrow V_{kh\text{í}}=V_{Cl_2}=0,115.22,4=2,576\left(l\right)\)

9. \(\left\{{}\begin{matrix}n_C=n_{CO_2}=\dfrac{9,9}{44}=0,225\left(mol\right)\\n_H=2n_{H_2O}=2.\dfrac{5,4}{18}=0,6\left(mol\right)\\n_{O\left(A\right)}=\dfrac{4,5-0,225.12-0,6}{16}=0,075\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_C:n_H:n_O=0,225:0,6:0,075=3:8:1\)

\(\Rightarrow A:\left(C_3H_8O\right)_n\)

\(\Rightarrow n=\dfrac{60}{60}=1\)

Vậy A là C₃H₈O

10) \(\left\{{}\begin{matrix}n_C=n_{CO_2}=\dfrac{22}{44}=0,5\left(mol\right)\\n_H=2n_{H_2O}=2.\dfrac{13,5}{18}=1,5\left(mol\right)\\n_O=\dfrac{7,5-0,5.12-1,5}{16}=0\left(mol\right)\end{matrix}\right.\)

``=> A`` không chứa O

\(\Rightarrow n_C:n_H=0,5:1,5=1:3\)

\(\Rightarrow A:\left(CH_3\right)_n\)

\(\Rightarrow n=\dfrac{15.2}{15}=2\)

Vậy A là C₂H₆

11) \(\left\{{}\begin{matrix}n_C=n_{CO_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\\n_H=2n_{H_2O}=2.\dfrac{0,18}{18}=0,02\left(mol\right)\\n_O=\dfrac{0,3-0,01.12-0,02}{16}=0,01\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_C:n_H:n_O=0,01:0,02:0,01=1:2:1\)

\(\Rightarrow A:\left(CH_2O\right)_n\)

\(\Rightarrow n=\dfrac{30.2}{30}=2\)

Vậy A là C₂H₄O₂

12) Gọi \(\left\{{}\begin{matrix}n_{CO_2}=4a\left(mol\right)\\n_{H_2O}=5a\left(mol\right)\end{matrix}\right.\)

Theo ĐLBTKL: \(44.4a+18.5a=2,25+32.\dfrac{3,08}{22,4}\Leftrightarrow a=0,025\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_C=n_{CO_2}=0,025.4=0,1\left(mol\right)\\n_H=2n_{H_2O}=5.2.0,025=0,25\left(mol\right)\\n_O=\dfrac{2,25-0,1.12-0,25}{16}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_C:n_H:n_O=0,1:0,25:0,05=2:5:1\)

\(\Rightarrow A:\left(C_2H_5O\right)_n\)

\(\Rightarrow n=\dfrac{45.2}{45}=2\)

Vậy A là C₄H₁₀O₂

13) Đề sai

a) \(n_{H_2SO_4}=\dfrac{5,88}{98}=0,06\left(mol\right)\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

            0,04<--0,06------->0,02---------->0,06

\(\Rightarrow\left\{{}\begin{matrix}a=m_{Al}=0,04.27=1,08\left(g\right)\\V=V_{H_2}=0,06.22,4=1,344\left(l\right)\end{matrix}\right.\)

b) 

Cách 1: \(m=m_{Al_2\left(SO_4\right)_3}=0,02.342=6,84\left(g\right)\)

Cách 2: \(m_{H_2}=0,06.2=0,12\left(g\right)\)

Áp dụng ĐLBTKL:

\(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)

\(\Rightarrow m=m_{Al_2\left(SO_4\right)_3}=1,08+5,88-0,12=6,84\left(g\right)\)

c) \(n_{O_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

PTHH: \(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,06}{2}< \dfrac{0,06}{1}\Rightarrow\) O₂ dư, H₂ hết

Theo PTHH: \(n_{O_2\left(p\text{ư}\right)}=\dfrac{1}{2}.n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(d\text{ư}\right)}=0,06-0,03=0,03\left(mol\right)\\m_{O_2\left(d\text{ư}\right)}=0,03.32=0,96\left(g\right)\\V_{O_2\left(d\text{ư}\right)}=0,03.22,4=0,672\left(l\right)\end{matrix}\right.\)

Theo PTHH: \(n_{H_2O}=n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,06.18=1,08\left(g\right)\)

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

Theo ĐLBTKL:
\(m_{t\text{ăn}g}=m_{O_2\left(p\text{ư}\right)}=3,2\left(g\right)\Rightarrow n_{O_2\left(p\text{ư}\right)}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

PTHH: \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

Theo PTHH: \(n_{Al\left(p\text{ư}\right)}=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,1=\dfrac{2}{15}\left(mol\right)< 0,5=n_{Al\left(b\text{đ}\right)}\)

`=>` Al dư, O₂ hết

\(n_{Al\left(d\text{ư}\right)}=0,5-\dfrac{2}{15}=\dfrac{11}{30}\left(mol\right)\)

Theo PTHH: \(n_{Al_2O_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,1=\dfrac{1}{15}\left(mol\right)\)

Vậy chất rắn sau phản ứng có: \(\left\{{}\begin{matrix}Al:m_{Al}=\dfrac{11}{30}.27=9,9\left(g\right)\\Al_2O_3:m_{Al_2O_3}=\dfrac{1}{15}.102=6,8\left(g\right)\end{matrix}\right.\)

a) \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PTHH: \(n_{Mg}=n_{MgCl_2}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m=m_{Mg}=0,5.24=12\left(g\right)\)

c) Theo PTHH: \(n_{HCl}=2n_{H_2}=1\left(mol\right)\)

\(\Rightarrow C_{M\left(HCl\right)}=\dfrac{1}{0,15}=\dfrac{20}{3}M\)

d) \(m_{ddHCl}=150.1=150\left(g\right)\)

\(\Rightarrow m_{dd.sau.p\text{ư}}=150+12-0,5.2=161\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,5.95}{161}.100\%=29,5\%\)

Ở \(90^oC:S_A=50\left(g\right)\)

\(\Rightarrow m_A=675.\dfrac{50}{100+50}=225\left(g\right)\)

\(\Rightarrow m_{H_2O}=675-225=450\left(g\right)\)

Ở \(20^oC:S_A=36\left(g\right)\)

\(\Rightarrow m_{A\left(tan.ra\right)}=\dfrac{450.36}{100}=162\left(g\right)\)

\(\Rightarrow m_{A\left(t\text{á}ch.ra\right)}=225-162=63\left(g\right)\)