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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Gọi hóa trị của kim loại M là n
M + nHCl → MCln + n/2H2
nHCl = = 0,6 mol
nM = => MM = = 12n
=> Với n = 2 và MM = 24 g/mol là giá trị thỏa mãn
Kim loại M là Magie (Mg)
\(a,X+2HCl\rightarrow XCl_2+H_2\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\\ n_X=n_{XCl_2}=n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ b,M_X=\dfrac{7,2}{0,3}=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:Magie\left(Mg=24\right)\\ c,m_{MgCl_2}=0,3.95=28,5\left(g\right)\\ d,V_{H_2\left(\text{đ}ktc\right)}=0,3.22,4=6,72\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
$a) 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$b) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$
Theo PTHH : $n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,6(mol)$
$m_{H_2SO_4} = 0,6.98 = 58,8(gam)$
$c) n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol) \Rightarrow m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)$
$d) n_{H_2} = n_{H_2SO_4} = 0,6(mol) \Rightarrow V_{H_2} = 0,6.22,4 = 13,44(lít)$
\(a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(b.n_{Al}=\dfrac{m}{M}=0,4\left(mol\right)\)
\(Theo.PTHH\Rightarrow n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}n_{Al}=1,5.0,4=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=n.M=0,6.98=58,8\left(g\right)\)
\(c,Theo.PTHH\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,5.0,4=0,2\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\\ d,V_{H_2\left(dktc\right)}=n.22,4=0,6.22,4=13,44\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 4X + 3O2 --to--> 2X2O3
2/15 <- 0,1 -------> 1/15
\(M_X=\dfrac{10,4}{\dfrac{2}{15}}=78\left(\dfrac{g}{mol}\right)\)
Bạn ơi đề có bị sai ko vậy :)?
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PT: \(4X+3O_2\underrightarrow{t^o}2X_2O_3\)
Theo PT: \(n_X=\dfrac{4}{3}n_{O_2}=0,2\left(mol\right)\)
\(\Rightarrow M_X=\dfrac{10,4}{0,2}=52\left(g/mol\right)\)
→ X là Crom.
b, \(n_{Cr_2O_3}=\dfrac{1}{2}n_{Cr}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cr_2O_3}=0,1.152=15,2\left(g\right)\)
c, \(Cr_2O_3+3H_2SO_4\rightarrow Cr_2\left(SO_4\right)_3+3H_2O\)
\(n_{H_2SO_4}=3n_{Cr_2O_3}=0,3\left(mol\right)\Rightarrow m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{H_2SO_4}=\dfrac{200.9,8}{100}=19,6\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
PTHH :
\(X+H_2SO_4\rightarrow XSO_4+H_2\)
0,2 0,2 0,2 0,2
\(M_X=\dfrac{8}{0,2}=40\left(dvC\right)\)
-> Canxi
\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(c,m_{CaSO_4}=0,2.136=27,2\left(g\right)\)
\(m_{ddCaSO_4}=8+200-\left(0,2.2\right)=207,6\left(g\right)\)
\(C\%=\dfrac{27,2}{207,6}.100\%\approx13,1\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ a,Fe+2HCl\rightarrow FeCl_2+H_2\\ b,n_{HCl}=0,4.2=0,8\left(mol\right)\\ m_{HCl}=0.8.36,5=29,2\left(g\right)\\ c,n_{H_2}=n_{FeCl_2}=n_{Fe}=0,4\left(mol\right)\\ m_{FeCl_2}=0,4.127=50,8\left(g\right)\\ d,V_{H_2\left(dktc\right)}=0,4.22,4=8,96\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
$X + 2HCl \to XCl_2 + H_2$
$2Y + 6HCl \to 2YCl_3 + 3H_2$
$n_{HCl} = \dfrac{47,45}{36,5} = 1,3(mol) \Rightarrow n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,65(mol)$
$\Rightarrow V_{H_2} = 0,65.22,4 = 14,56(lít)$
b) Bảo toàn khối lượng : $m_{muối} = 12,9 + 1,3.36,5 - 0,65.2 = 59,05(gam)$
c) Gọi $n_X = a(mol) \Rightarrow n_{Al} = 1,5a(mol)$
Theo PTHH : $n_{H_2} = a + 1,5a.\dfrac{3}{2} = 0,65(mol) \Rightarrow a = 0,2$
$\Rightarrow m_{hh} = 0,2.X + 0,2.1,5.27 = 12,9$
$\Rightarrow X = 24(Magie)$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{1,625}{65}=0,025mol\)
\(n_{HCl}=\dfrac{3,65}{36,5}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,025 < 0,1 ( mol )
0,025 0,05 0,025 0,025 ( mol )
\(V_{H_2}=0,025.22,4=0,56l\)
Chất dư là HCl
\(m_{HCl\left(dư\right)}=\left(0,1-0,05\right).36,5=1,825g\)
\(m_{ZnCl_2}=0,025.136=3,4g\)
huhu cảm ơn bạn đề dài quá nên đưa lên đây làm giúp
![](https://rs.olm.vn/images/avt/0.png?1311)
4) x,y lần lượt là số mol của M và M2O3
=> nOxi=3y=nCO2=0,3 => y=0,1
Đề cho x=y=0,1 =>0,1M+0,1(2M+48)=21,6 =>M=56 => Fe và Fe2O3
=> m=0,1.56 + 0,1.2.56=16,8
2)X + 2HCl === XCl2 + H2
n_h2 = 0,4 => X = 9,6/0,4 = 24 (Mg)
=>V_HCl = 0,4.2/1 = 0,8 l
\(a.PTHH:2X+3H_2SO_4--->X_2\left(SO_4\right)_3+3H_2\uparrow\)
b. Ta có: \(n_{H_2SO_4}=\dfrac{17,64}{98}=0,18\left(mol\right)\)
Theo PT: \(n_X=\dfrac{2}{3}.n_{H_2SO_4}=\dfrac{2}{3}.0,18=0,12\left(mol\right)\)
\(\Rightarrow M_X=\dfrac{3,24}{0,12}=27\left(\dfrac{g}{mol}\right)\)
Vậy X là kim loại nhôm (Al)
\(c.PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,12=0,06\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,06.342=20,52\left(g\right)\)
d. Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,18\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,18.22,4=4,032\left(lít\right)\)