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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: \(2A+3H_2SO_4\rightarrow A_2\left(SO_4\right)_3+3H_2\)
=> \(n_A=0,4\left(mol\right)\)
=> \(M_A=\dfrac{10,8}{0,4}=27\left(g/mol\right)\)
=> A là Al
b) \(n_{Al_2\left(SO_4\right)_3}=0,2\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có: \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,3}{3}\), ta được Fe2O3 dư.
Theo PT: \(n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\Rightarrow n_{Fe_2O_3\left(dư\right)}=0,3-0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,2.160=32\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(nZn=\dfrac{13}{65}=0,2\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
1 1 1 1 (mol)
0,2 0,2 0,2 0,2 (mol)
\(VH_2=0,2.22,4=4,48\left(l\right)\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
1 3 2 3 (mol)
0,2 2/15 (mol)
\(mFe=\dfrac{2}{15}.56=7,47\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nH2 = 1,344 : 22,4 = 0,06(mol)
pthh 2M+ 3H2SO4 ---> M2(SO4)3+ 3H2
0,04<-- 0,06---------------------------0,06(mol)
M M = 1,08 : 0,04 = 27 (g/mol )
=> M : Al
mH2SO4 = 0,06.98 =5,88 (g)
nH2 = 1,344/22,4 = 0,06 (mol)
PTHH:
2M + 3H2SO4 -> Al2(SO4)3 + 3H2
0,04 <--- 0,06 <--- 0,02 <--- 0,06
M(M) = 1,08/0,04 = 27 (g/mol(
=> M là Al
mH2SO4 = 0,06 . 98 = 5,88 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2M+2nHCl\rightarrow2MCl_n+nH_2\)
\(n_M=\dfrac{2}{n}n_{H_2}=\dfrac{0,5}{n}\left(mol\right)\)
Ta có : \(M_M=\dfrac{16,25}{\dfrac{0,5}{n}}=32,5n\)
Chạy nghiệm n
n=1 => M=32,5 (loại)
n=2 => M=65 ( chọn)
n=3 => M=97,5 (loại)
Vậy M là Zn
b) Ta có : \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)
=> \(V_{HCl}=\dfrac{0,5}{0,2}=2,5\left(lít\right)\)
\(2A+3H_2SO_4\rightarrow A_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\Rightarrow n_A=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ a,M_A=\dfrac{5,4}{0,2}=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Nhôm\left(Al=27\right)\\ b,n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,3}{0,2}=1,5\left(lít\right)\\ c,\\ 3H_2+Fe_2O_3\rightarrow\left(t^o\right)2Fe+3H_2O\\ n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\\ n_{H_2}=0,3\left(mol\right)\\ Vì:\dfrac{0,3}{3}< \dfrac{0,2}{1}\Rightarrow Fe_3O_4dư\\ n_{Fe}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ n_{Fe_3O_4\left(dư\right)}=0,2-\dfrac{1}{3}.0,3=0,1\left(mol\right)\\ m_{rắn}=m_{Fe_3O_4\left(dư\right)}+m_{Fe}=0,1.232+0,2.56=34,4\left(g\right)\)
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