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![](https://rs.olm.vn/images/avt/0.png?1311)
Tham khảo:
2A+ 2H2O -> 2AOH+ H2
- TN1:
nH2= 0,05 mol
nA= 2,1/A mol
=> 1,05/A < 0,05
=> A > 21 (1)
- TN2:
nH2= 0,1 mol
nA= 8,2/A mol
=> 4,1/A > 0,1
=> A < 41 (2)
(1)(2) => 21 < A < 41
Vậy A= 23 (Na) hoặc A= 39 (K)
![](https://rs.olm.vn/images/avt/0.png?1311)
2A + H20 ---> H2 + A20
Trường hợp 1 : nH2 < 0,05 ⇒⇒ nA < 0,1 => MA > 2,1 / 0,1 = 21
Trường hợp 2 : nH2 > 0,1 => nA > 0,2 => MA < 41
=> A là Rb ( 37 , hóa trị l )
![](https://rs.olm.vn/images/avt/0.png?1311)
2A + H2O \(\rightarrow\) H2 + A2O
Trường hợp 1 : nH2 < 0,05 \(\Rightarrow\) nA < 0,1 \(\Rightarrow\) MA > 2,1/0,1 = 21
Trường hợp 2 : nH2 > 0,1 \(\Rightarrow\) nA > 0,2 \(\Rightarrow\) MA < 41
\(\Rightarrow\) A là Rb ( 37 , hóa trị 1 )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{A\left(1\right)}=\frac{2,1}{A}\left(mol\right);n_{A\left(2\right)}=\frac{8,2}{A}\left(mol\right)\)
PTHH 1: 2A + 2H2O --> 2AOH + H2
\(\frac{2,1}{A}\) -------------------------> \(\frac{1,05}{A}\) (mol)
PTHH 2: 2A + 2H2O --> 2AOH + H2
\(\frac{8,2}{A}\) --------------------------> \(\frac{4,1}{A}\) (mol)
=> \(\frac{1,05}{A}< \frac{1,12}{22,4}\) ; \(\frac{4,1}{A}>\frac{2,24}{22,4}\)
=> \(\frac{1,05}{A}< 0,05\); \(\frac{4,1}{A}>0,1\)
=> 21<A<41 => A là Na,K
MIK NGHĨ ZẬY
\(2A+2H_2O\rightarrow2Aoh+H_2\)
- TN1:
\(n_{H2}=0,05\left(mol\right)\)
\(n_A=\frac{2,1}{A}\left(mol\right)\)
\(\Rightarrow\frac{1,05}{A}< 0,05\Rightarrow A>21\left(1\right)\)
- TN2:
\(n_{H2}=0,1\left(mol\right)\)
\(n_A=\frac{8,2}{A}\left(mol\right)\)
\(\Rightarrow\frac{4,1}{A}>0,1\Rightarrow A< 41\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow21< A< 41\)
Vậy A= 23 (Na) hoặc A= 39 (K)
![](https://rs.olm.vn/images/avt/0.png?1311)
2) Gọi kim loại hóa trị II là x
X + 2H2O → X(OH)2 + H2
nH2 = 2,24:22,4 =0,1 mol
nX = \(\dfrac{4}{^MX}\)=nH2
=> \(\dfrac{4}{^MX}\)=0,1 => MX=40 => X là kim loại Canxi (Ca)
Bài 1:
a, Ta có: \(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\)
\(n_{H_2O}=\dfrac{45}{18}=2,5\left(mol\right)\)
PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{2,5}{3}\), ta được H2O dư.
Theo PT: \(n_{H_2O\left(pư\right)}=3n_{P_2O_5}=0,3\left(mol\right)\)
\(\Rightarrow n_{H_2O\left(dư\right)}=2,5-0,3=2,2\left(mol\right)\)
\(\Rightarrow m_{H_2O\left(dư\right)}=2,2.18=39,6\left(g\right)\)
b, Theo PT: \(n_{H_3PO_4}=2n_{P_2O_5}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_3PO_4}=0,2.98=19,6\left(g\right)\)
Bạn tham khảo nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 24:
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:A+2HCl\rightarrow ACl_2+H_2\uparrow\)
Theo pthh: nA = nH2 = 0,15 (mol)
=> MA = \(\dfrac{3,6}{0,15}=24\left(\dfrac{g}{mol}\right)\)
=> A là Mg
Bài 25:
\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ PTHH:2A+6HCl\rightarrow2ACl_3+3H_2\uparrow\\ Mol:0,3\leftarrow0,9\leftarrow0,3\leftarrow0,45\\ \rightarrow\left\{{}\begin{matrix}M_A=\dfrac{8,1}{0,3}=27\left(\dfrac{g}{mol}\right)\Rightarrow A:Al\\m_{HCl}=0,9.36,5=32,85\left(g\right)\end{matrix}\right.\)
Bài 24.
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_A=\dfrac{3,6}{M_A}\) mol
\(A+2HCl\rightarrow ACl_2+H_2\)
0,15 0,15 ( mol )
\(\Rightarrow\dfrac{3,6}{M_A}=0,15mol\)
\(\Leftrightarrow M_A=24\) ( g/mol )
=> A là Magie ( Mg )
Bài 25.
\(n_{H_2}=\dfrac{10,08}{22,4}=0,45mol\)
\(n_A=\dfrac{8,1}{M_A}\) mol
\(2A+6HCl\rightarrow2ACl_3+3H_2\)
0,3 0,45 ( mol )
\(\Rightarrow\dfrac{8,1}{M_A}=0,3\)
\(\Leftrightarrow M_A=27\) g/mol
=> A là nhôm ( Al )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2A+3H_2SO_4\rightarrow A_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\Rightarrow n_A=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ a,M_A=\dfrac{5,4}{0,2}=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Nhôm\left(Al=27\right)\\ b,n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,3}{0,2}=1,5\left(lít\right)\\ c,\\ 3H_2+Fe_2O_3\rightarrow\left(t^o\right)2Fe+3H_2O\\ n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\\ n_{H_2}=0,3\left(mol\right)\\ Vì:\dfrac{0,3}{3}< \dfrac{0,2}{1}\Rightarrow Fe_3O_4dư\\ n_{Fe}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ n_{Fe_3O_4\left(dư\right)}=0,2-\dfrac{1}{3}.0,3=0,1\left(mol\right)\\ m_{rắn}=m_{Fe_3O_4\left(dư\right)}+m_{Fe}=0,1.232+0,2.56=34,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2R + 2H_2O \to 2ROH + H_2\\ n_{H_2} > \dfrac{7,5}{22,4} = \dfrac{75}{224}\\ \Rightarrow n_R = 2n_{H_2} > \dfrac{75}{112}\\ \Rightarrow M_R < \dfrac{4,9}{\dfrac{75}{112}} = 7,3\\ \Rightarrow M_R = 7(Li)\)
Vậy kim loại R là Liti
\(2A+2HCl\rightarrow2ACl+H_2\)
TH1:
\(n_A=\frac{2,1}{A}\left(mol\right)\)
\(\Rightarrow n_{H_2}=\frac{2,1}{2A}< \frac{1,12}{22,4}=0,05\)
\(\Leftrightarrow A>21\left(1\right)\)
TH2:
\(n_A=\frac{8,2}{A}\left(mol\right)\)
\(\Rightarrow n_{H_2}=\frac{8,2}{2A}>\frac{2,24}{22,4}=0,1\)
\(\Leftrightarrow A< 41\left(2\right)\)
Từ (1) và (2) ta suy ra A là Na hoặc K
Hi