\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

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

0,2-->0,4----->0,2------->0,2

a

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b

\(CM_{MgCl_2}=\dfrac{0,2}{0,2}=1M\)

c

\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)

0,2------>0,4

\(V_{dd.NaOH}=\dfrac{0,4}{2}=0,2\left(l\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(m_{HCl}=21,9g\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\)

=> HCl dư

\(\Rightarrow n_{H_2}=0,2mol\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48l\)

bổ sung ý b)

Khối lượng dung dịch sau phản ứng = mZn + mHCl - mH₂ thoát ra = 13 +150 - 0,2 .2 = 162,6 gam

Dung dịch thu được sau phản ứng gồm \(\left\{{}\begin{matrix}ZnCl_2\\HCl_{dư}\end{matrix}\right.\)

nZnCl₂ = nZn = 0,2 mol => mZnCl₂ = 0,2 . 136 = 27,2 gam

=> C% ZnCl₂ = \(\dfrac{27,2}{162,6}\).100= 16,72%

nHCl dư = 0,6 - 0,4 = 0,2 mol 

mHCl dư= 0,2.36,5 = 7,3 gam

=> C% HCl dư = \(\dfrac{7,3}{162,6}\).100 = 4,5%

a, PT: \(Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)

Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=n_{Na_2SO_3}=0,1\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(n_{NaCl}=n_{HCl}=2n_{Na_2SO_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

c, \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{7,3}{10\%}=73\left(g\right)\)

d, Ta có: m _{dd sau pư} = 12,6 + 73 - 0,1.64 = 79,2 (g)

\(\Rightarrow C\%_{NaCl}=\dfrac{11,7}{79,2}.100\%\approx14,77\%\)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

⇒ m_FeO = 12,6 - 5,4 = 7,2 (g)

c, Phần này đề cho dd NaOH dư hay vừa đủ bạn nhỉ?

d, Cho hh vào dd H₂SO₄ đặc nguội thì có khí thoát ra.

PT: \(2FeO+4H_2SO_{4\left(đ\right)}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)

Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=\dfrac{1}{2}n_{FeO}=0,05\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

gfvfvfvfvfvfvfv555

`1)`

`n_{Al}={2,7}/{27}=0,1(mol)`

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

`0,1->0,15->0,05->0,15(mol)`

`V_{dd\ H_2SO_4}={0,15}/1=0,15(l)=150(ml)`

`->V=150`

`V'=V_{H_2}=0,15.22,4=3,36(l)`

`C_{M\ X}=C_{M\ Al_2(SO_4)_3}={0,05}/{0,15}=1/3M`

`2)`

`n_{Fe}={2,8}/{56}=0,05(mol)`

`Fe+2HCl->FeCl_2+H_2`

`0,05->0,1->0,05->0,05(mol)`

`V_{dd\ HCl}={0,1}/1=0,1(l)=100(ml)`

`->V=100`

`V_{H_2}=0,05.22,4=1,12(l)`

`C_{M\ FeCl_2}={0,05}/{0,1}=0,5M`

\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

a) Chất tan : FeSO₄

Chất khí : H₂

\(m_{FeSO_4}=0.05\cdot152=7.6\left(g\right)\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

Fe + H2SO4 -> FeSO4 + H2

 1           1                 1            1

A là: FeSO4

B là: H2

nFe = 2,8/56 = 0,05 mol => nH2SO4 = 0,05 mol

mFeSO4 = 0,05 x (56+32+64) = 7,6 gam 

VH2 = 0,05 x 22,4 = 1,12 lit

a) Fe+2HCl--->FeCl2+H2

n Fe=2,8/56=0,05(mol)

n HCl=2n Fe=0,1(mol)

V HCl=0,1/2=0,05(l)

b) n H2=n Fe=0,05(mol)

V H2=0,05.22,4=1,12(l)

c) n FeCl2=n Fe=0,05(mol)

Do thể tích dd thay đổi k đáng kể nên V dd=V HCl=0,05(l)

CM FeCl2=0,05/0,05=1(M)

a) PT Fe+2HCl--->FeCl2+H2

n Fe=2,8/56=0,05(mol)

ta có :n HCl=2n Fe=0,1(mol)

=>V HCl=0,1/2=0,05(l)

b) ta có : n H2=n Fe=0,05(mol)

=>V H2=0,05.22,4=1,12(l)

c) n FeCl2=n Fe=0,05(mol)

=> V dd=V HCl=0,05(l)

CM _FeCl2=0,05/0,05=1(M)

Bài 1: Ta có: \(n_{H_2}=\dfrac{1,008}{22,4}=0,045\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

__0,045__0,09____0,045___0,045 (mol)

a, Ta có: \(a=m_{Mg}=0,045.24=1,08\left(g\right)\)

b, \(V_{ddHCl}=\dfrac{0,09}{0,1}=0,9\left(l\right)\)

c, \(C_{M_{MgCl_2}}=\dfrac{0,045}{0,9}=0,05M\)

Bài 2:

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 56y = 5,2 (1)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=x+y\left(mol\right)\)

⇒ x + y = 0,15 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{5,2}.100\%\approx46,2\%\\\%m_{Fe}\approx53,8\%\end{matrix}\right.\)

b, Ta có: \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,3}{1}=0,3\left(l\right)\)

Bạn tham khảo nhé!