2. \(1.n_{H_2}=0,1\left(mol\right);n_{CuO}=0,15\left(mol\right)\\ H_2+CuO-^{t^o}\rightarrow Cu+H_2O\\ LTL:\dfrac{0,1}{1}< \dfrac{0,15}{1}\Rightarrow CuOdư\\ n_{CuO\left(pư\right)}=n_{H_2}=0,1\left(mol\right)\\ n_{CuO\left(dư\right)}=0,15-0,1=0,05\left(mol\right)\\ \Rightarrow m_{CuO\left(dư\right)}=4\left(g\right)\\ 2.n_{H_@}=n_{Cu}=0,1\left(mol\right)\\ \Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

Bài 9:

Gọi hóa trị của M là x(x>0)

\(n_{H_2}=\dfrac{1,008}{22,4}=0,045(mol)\\ \text {Bảo toàn O: }n_{O/oxit}=n_{H_2}=\dfrac{1,344}{22,4}=0,06(mol)\\ \Rightarrow m_{M}=m_{oxit}-m_{O_2}=3,48-0,06.16=2,52(g)\\ PTHH:2M+2xHCl\to 2MCl_x+xH_2\\ \Rightarrow n_{M}=\dfrac{2.0,045}{x}=\dfrac{0,09}{x}\\ \Rightarrow M_{M}=\dfrac{2,52}{\dfrac{0,09}{x}}=28x\)

Thay \(x=2\Rightarrow M_M=56(Fe)\)

\(\Rightarrow n_{Fe}=\dfrac{2,52}{56}=0,045(mol)\\ \Rightarrow \dfrac{n_{Fe}}{n_{O}}=\dfrac{0,045}{0,06}=\dfrac{3}{4}\)

Vậy M là sắt (Fe) và CTHH oxit là \(Fe_3O_4\)

9. Tham khảo

Bài 12:

\(\text{Đ}\text{ặt}:X\left(a\right)\left(x:nguy\text{ê}n,d\text{ư}\text{ơn}g\right)\\ 2X+2aHCl\rightarrow2XCl_a+H_2\\ n_{H_2}=\dfrac{4,704}{22,4}=0,21\left(mol\right)\\ \Rightarrow n_X=0,21.2=0,42\left(mol\right)\\ M_X=\dfrac{3,78}{0,42}=9\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:B\text{er}i\left(Be=9\right)\)

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Bài 11:

\(m_{r\text{ắn}\left(gi\text{ảm}\right)}=m_{O\left(trong.\text{ox}it\right)}=4,8\left(g\right)\\ \Rightarrow n_{O\left(trong.\text{ox}it\right)}=\dfrac{4,8}{16}=0,3\left(mol\right)\\ n_{Fe}=\dfrac{16-4,8}{56}=0,2\left(mol\right)\\ \text{Đ}\text{ặt}.CTTQ:Fe_xO_y\left(x,y:nguy\text{ên},d\text{ư}\text{ơ}ng\right)\\ x:y=0,2:0,3=2:3\\ \Rightarrow x=0,2;y=0,3\\ \Rightarrow CTHH:Fe_2O_3\)

a) n Fe = 5,6/56 = 0,1(mol) ; n HCl = 10,95/36,5 = 0,3(mol)

$Fe + 2HCl \to FeCl_2 + H_2$
Ta thấy : 

n Fe / 1 = 0,1 < n HCl / 2 = 0,3/2 = 0,15 nên HCl dư

Theo PTHH : n HCl pư = 2n Fe = 0,2(mol)

Suy ra: m HCl dư = 10,95 - 0,2.36,5 = 3,65(gam)

b)

Theo PTHH : n FeCl2 = n H2 = n Fe = 0,1(mol)

m FeCl2 = 0,1.127 = 12,7 (gam)

m H2 = 0,1.2 = 0,2(gam)

c)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

n Cu = n H2 = 0,1(mol)

m Cu = 0,1.64 = 6,4(gam)

a) Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

b)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{4}\) => H₂ hết, Fe₃O₄ dư

PTHH: Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

             0,025<--0,1------>0,075

=> \(m_{Fe_3O_4\left(dư\right)}=\left(0,1-0,025\right).232=17,4\left(g\right)\)

c) \(m_{Fe}=0,075.56=4,2\left(g\right)\)

ai giúp mik câu b với ạ

HgO+H₂-to>Hg+H₂O

0,025-0,025--0,025

n _HgO=\(\dfrac{8,68}{217}\)=0,04 mol

n _H2=\(\dfrac{0,56}{22,4}\)=0,025 mol

=>HgO dư

=>m Hg=0,025.201=5,025g

nH2 = 0,56/22,4 = 0,025 (mol)

nHgO = 8,68/217 = 0,04 (mol)

PTHH: HgO + H2 -> (t°) Hg + H2O

LTL: 0,025 < 0,04 => H2 dư

nH2 (p/ư) = nHg = 0,025 (mol)

VH2 = (0,04 - 0,025) . 22,4 = 0,336 (l)

mHg = 0,025 . 201 = 5,025 (g)

Ta có: \(n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\)
PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT ta có: \(n_{CuO}=n_{Cu}=0,025\left(mol\right)\)
Khối lượng đồng (II) oxit phản ứng là:
\(m_{CuO}=0,025.80=2\left(g\right)\)

\(n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\)

PTHH: \(H_2+CuO\underrightarrow{t^0}Cu+H_2O\)

Theo PTHH: \(n_{CuO}:n_{Cu}=1:1\Rightarrow n_{CuO}:n_{Cu}=0,025\left(mol\right)\)

\(\Rightarrow m_{CuO}=n_{CuO}.M_{CuO}=0,025.80=2\left(g\right)\)

PTHH. CuO + H2 -> H2O + Cu

Theo bài:
nCu = \(\dfrac{1,6}{64}=0,025\left(mol\right)\)

Theo pthh và bài có:

nCuO = nCu = 0,025 (mol)

=>mCuO = 0,025 . 80 = 2 (g)

a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

           0,1---->0,1------->0,1---->0,1

=> \(m_{dd.H_2SO_4}=\dfrac{0,1.98}{4,9\%}=200\left(g\right)\)

b) m_{dd sau pư} = 2,4 + 200 - 0,1.2 = 202,2 (g)

m_MgSO4 = 0,1.120 = 12 (g)

\(C\%_{MgSO_4}=\dfrac{12}{202,2}.100\%=5,9\%\)

c) 

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,25}{1}>\dfrac{0,1}{1}\) => Hiệu suất tính theo H₂

\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

                      0,05<-----0,05

=> \(H=\dfrac{0,05}{0,1}.100\%=50\%\)

a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH: Mg + H₂SO₄ ---> MgSO₄ + H₂

           0,1--->0,1---------->0,1-------->0,1

\(m_{dd\left(H_2SO_4\right)}=\dfrac{0,1.98}{4,9\%}=200\left(g\right)\)

b, \(m_{dd\left(sau.pư\right)}=2,4+200-0,2.2=202,2\left(g\right)\)

\(\rightarrow C\%_{MgSO_4}=\dfrac{0,1.120}{202,2}.100\%=5,93\%\)

c, \(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

LTL: 0,25 > 0,1 => CuO dư

\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

Theo pt: \(n_{H_2}=n_{Cu}=0,05\left(mol\right)\)

=> \(H=\dfrac{0,05}{0,1}.100\%=50\%\)