\(n_{O_2\left(pứ\right)}=\dfrac{9,3-6,9}{32}=0,075\left(mol\right)\)

Gọi kim loại hóa trị I cần tìm là R.

PTHH:

\(4R+O_2\underrightarrow{t^o}2R_2O\)

\(n_R=0,075.4=0,3\left(mol\right)\\ \Rightarrow M_R=\dfrac{6,9}{0,3}=23\)

Vậy tên kim loại là Na

Gọi kim loại cần tìm là `M`

`4M + O_2 overset(t^o)(->) 2M_2O`

Theo PT `:n_(M) =2 n_(M_2O)`

`<=> (6,9)/(M) = 2 . (9,3)/(2M + 16)`

Giải ra được `:`

`M_(M)=23(g//mol)`

`=>M` là `Na`

các bạn giúp mình nha

\(1.a.CTHH:M_2O_3\\ 4M+3O_2\xrightarrow[]{t^0}2M_2O_3\\ \Rightarrow n_M:4=n_{M_2O_3}:2\\ \Rightarrow\dfrac{2,7}{M}:4=\dfrac{5,1}{2M+48}:2\\ \Rightarrow M=27,Al\\ \Rightarrow CTHH:Al_2O_3\\ bAl_2O_3:Alumium.oxide\\ 2.KL:R\\ R+2HCl\rightarrow RCl_2+H_2\\ n_R=\dfrac{1}{2}n_{HCl}=0,1mol\\ M_R=\dfrac{2,4}{0,1}=24\\ M.là.Mg\)

\(n_{Al}=\dfrac{10,8}{27}=0,4mol\\ 2Al+3S\xrightarrow[t^0]{}Al_2S_3\\ n_{Al_2S_3\left(lt\right)}=\dfrac{1}{2}n_{Al}=0,2mol\\ m_{Al_2S_3\left(lt\right)}=0,2.150=30g\\ H=\dfrac{25,5}{30}\cdot100\%=85\%\)
 

1.

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

2. Bạn coi lại đề

3.

\(a)n_C=\dfrac{3}{12}=0,25mol\\ C+O_2\xrightarrow[]{t^0}CO_2\\ n_{O_2}=n_C=0,25mol\\ V_{CO_2}=0,25.24,79=6,1975l\\ b)n_O=2n_C=0,5mol\\ V_{CO_2}=0,5.24,79=12,395l\)

a) dd HCl 

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

\(Mg\left(OH\right)_2+2HCl\rightarrow MgCl_2+2H_2O\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

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

b) dd H₂SO₄

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(Mg\left(OH\right)_2+H_2SO_4\rightarrow MgSO_4+2H_2O\)

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

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

a b thiếu Zn:)

Gọi CTHH của Caffein là \(C_xH_yO_zN_a\)

\(\%N=100\%-48,98\%-6,12\%-16,33\%=28,57\%\)

Ta có tỉ lệ:

\(\dfrac{12x}{48,98}=\dfrac{y}{6,12}=\dfrac{16z}{16,33}=\dfrac{14a}{28,57}=\dfrac{196}{100}\\ \Leftrightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=2\\a=4\end{matrix}\right.\)

Vậy CTHH của caffein là \(C_8H_{12}O_2N_4\)

\(n_{H_2SO_4}=0,6.1=0,6mol\\ n_{ZnO}=a;n_{Fe_2O_3}=b\\ a.ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ b.\Rightarrow\left\{{}\begin{matrix}81a+160b=40,3\\a+3b=0,6\end{matrix}\right.\\ \Rightarrow a=0,3;b=0,1\\ \%m_{ZnO}=\dfrac{0,3.81}{40,3}\cdot100\%=60,3\%\\ \%m_{Fe_2O_3}=100\%-60,3\%=39,7\%\\ c.n_{ZnSO_4}=n_{Zn}=0,3mol\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1mol\\ C_{M_{ZnSO_4}}=\dfrac{0,3}{0,6}=0,5M\\ C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,6}M\)

\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ n_{Mg}=a,n_{Na}=b\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow\left\{{}\begin{matrix}24a+23b=9,4\\a+0,5b=0,3\end{matrix}\right.\\ \Rightarrow a=b=0,2\\ m_{Mg}=0,2.24=4,8g\\ m_{Na}=0,2.23=3,6g\\ b.n_{H_2SO_4}=0,5b+a=0,3mol\\ m_{ddH_2SO_4}=\dfrac{0,3.98}{10\%}\cdot100\%=294g\\ c.n_{MgSO_4}=n_{Mg}=0,2mol\\ n_{Na_2SO_4}=0,5n_{Na}=0,1mol\\ m_{MgSO_4}=120.0,2=24g\\ m_{Na_2SO_4}=142.0,1=14,2g\)