Bài 1 :

PTHH : Zn + H₂SO₄ ------> ZnSO₄ + H₂

\(n_{Zn}=\frac{m}{M}=\frac{26}{65}=0,4\left(mol\right)\)

Theo PTHH : n_H2 = n_Zn = 0,4 mol

=> Khối lượng H₂ được tạo ra ở đktc là :

\(V=n\times22,4\)

\(\Rightarrow V_{H_2}=0,4\times22,4\)

\(\Rightarrow V_{H_2}=8,96\left(l\right)\)

Theo PTHH : n_ZnSO4 = n_Zn = 0,4 mol

=> Khối lượng muối được tạo thành là :

\(m=n\times M\)

\(\Rightarrow m_{ZnSO_4}=0,4\times161\)

\(\Rightarrow m_{ZnSO_4}=64,4\left(g\right)\)

b) Theo PT : \(n_{H_2SO_4}=n_{Zn}=0,4\left(mol\right)\)

=> Khối lượng \(H_2SO_4\)cần dùng cho phản ứng là :

\(m=n\times M\)

\(\Rightarrow m_{H_2SO_4}=0,4\times98\)

\(\Rightarrow m_{H_2SO_4}=39,2\left(g\right)\)

c) Nồng độ phần trăm thu được sau phản ứng là :

\(C\%=\frac{m_{ct}}{m_{dd}}\times100\%\)

\(\Rightarrow C\%=\frac{39,2}{64,4}\times100\%\approx60,9\%\)

Vậy :.........................

\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)

        2             3                  1                3

       0,8          1,2               0,4             1,2

a) \(n_{H2}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=1,2.22,4=26,88\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

⇒ \(m_{H2SO4}=1,2.98=117,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{117,6.100}{29,4}=400\left(g\right)\)

c) \(n_{Al2\left(SO4\right)3}=\dfrac{1,2.1}{3}=0,4\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,4.342=136,8\left(g\right)\)

\(m_{ddspu}=21,6+400-\left(1,2.2\right)=419,2\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{136,8.100}{419,2}=32,63\)⁰/₀

 Chúc bạn học tốt

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

b, Ta có: \(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{1,47}{98}=0,015\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,015}{1}\), ta được Zn dư.

Theo PT: \(n_{Zn\left(pư\right)}=n_{H_2SO_4}=0,015\left(mol\right)\Rightarrow n_{Zn\left(dư\right)}=0,03-0,015=0,015\left(mol\right)\)

\(\Rightarrow m_{Zn\left(dư\right)}=0,015.65=0,975\left(g\right)\)

c, \(n_{H_2}=n_{H_2SO_4}=0,015\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,015.22,4=0,336\left(l\right)\)

1////          nAl=0,4mol   

2Al     +      6HCl -----> 2AlCl3 + 3H2

0,4mol      1,2mol         0,4mol    0,6 mol

a/ V_H2=0,6.22,4=13,44 l

b/ V=1,2/2=0,6 l

C_AlCl3=0,4/0,6=2/3 M

a)

$n_{Zn} = \dfrac{13}{65} = 0,2(mol) ; n_{H_2 SO_4} = 0,5.2 = 1(mol)$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Ta thấy : 

$n_{Zn} < n_{H_2SO_4}$ nên $H_2SO_4$ dư

$n_{ZnSO_4} = n_{H_2SO_4\ pư} = n_{Zn} = 0,2(mol)$
$m_{ZnSO_4} = 0,2.161=32,2(gam)$
$m_{H_2SO_4\ pư} = 0,2.98 = 19,6(gam)$

b)

$n_{H_2SO_4\ dư} = 1 - 0,2 = 0,8(mol)$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,8}{0,5} = 1,6M$
$C_{M_{FeSO_4}} = \dfrac{0,2}{0,5} = 0,4M$

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,5.2=1\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ a.Vì:\dfrac{0,2}{1}< \dfrac{1}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{Zn}=0,2\left(mol\right)\\ m_{H_2SO_4\left(p.ứ\right)}=0,2.98=19,6\left(g\right)\\ m_{ZnSO_4}=161.0,2=32,2\left(g\right)\\ b.V_{ddsau}=V_{ddH_2SO_4}=0,5\left(l\right)\\ C_{MddZnSO_4}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\ C_{MddH_2SO_4\left(dư\right)}=\dfrac{1-0,2}{0,5}=1,6\left(M\right)\)

a) Zn + H₂SO₄ --> ZnSO₄ + H₂

b) \(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{1,47}{98}=0,015\left(mol\right)\)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,015}{1}\) => Zn dư, H₂SO₄ hết

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

____0,015<-0,015--->0,015->0,015

=> m_Zn(dư) = (0,03-0,015).65 = 0,975 (g)

c) V_H2 = 0,015.22,4 = 0,336(l)

Chào bạn, Uyển Lộc 🤗

a)\(PTHH:Zn+H_2SO_4\underrightarrow{ }ZnSO_4+H_2\)

b)\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(m\right)\);\(n_{H_2SO_4}=\dfrac{1,57}{98}=0,16\left(m\right)\)

\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

ta có tỉ lệ:\(\dfrac{0,3}{1}>\dfrac{0,16}{1}->Zndư\)

\(n_{Zn\left(dư\right)}=0,3-0,16=0,14\left(m\right)\)

\(m_{Zn\left(dư\right)}=0,14.65=9,1\left(g\right)\)

c)\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

tỉ lệ        :1       1                1              1

số mol   :0,16   0,16           0,16         0,16

\(V_{H_2}=0,16.22,4=3,584\left(l\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{1,57}{98}\approx0,016\left(mol\right)\)

\(PT:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(\dfrac{n_{Zn\left(ĐB\right)}}{n_{Zn\left(PT\right)}}=\dfrac{0,03}{1}>\dfrac{n_{H_2SO_4\left(ĐB\right)}}{n_{H_2SO_4}\left(PT\right)}=\dfrac{0,016}{1}\)

\(\Rightarrow\) Zn dư , H₂SO₄ hết , tính theo H₂SO₄  

b, Theo PT : \(n_{zn}=n_{H_2SO_4}=0,016\left(mol\right)\)

\(\Rightarrow m_{Zn\left(pứ\right)}=n\cdot M=0,016\cdot32=0,512\left(g\right)\)

\(\Rightarrow m_{Zn\left(dư\right)}=m_{Zn\left(ĐB\right)}-n_{Zn\left(Pứ\right)}=1,95-0,512=1,438\left(g\right)\)

c, Theo PT : \(n_{H_2}=n_{H_2SO_4}=0,016\left(mol\right)\)

\(\Rightarrow V_{H_{2\left(đktc\right)}}=n\cdot22,4=0,016\cdot22,4=0,3584\left(l\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Theo bài ra ta có :

\(m_{Zn}=1,95\left(g\right)\Rightarrow n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(m_{H_2SO_4}=1,57\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{m_{H_2SO_4}}{M_{H_2SO_4}}=\dfrac{1,57}{98}\approx0,016\left(mol\right)\)

Ta có tỉ lệ số mol chia cho hệ số :

\(\dfrac{0,03}{1}>\dfrac{0,016}{1}\)

\(\Rightarrow\) H2SO4 phản ứng hết / Zn phản ứng dư

\(\Rightarrow\) Tính theo số mol của H2SO4

⇒ \(n_{Zn}\) tham gia phản ứng là : \(n_{Zn}\approx0,016\left(mol\right)\Rightarrow m_{Zn}\approx0,016.32\approx0,512\left(g\right)\Rightarrow m_{Zn}\left(dư\right)=1,438\left(g\right)\)

\(V_{H_2}=0,016.22,4\approx0,3584\left(l\right)\)

PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

a, Ta có: \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=\dfrac{200.20}{100}=40\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{40}{98}=\dfrac{20}{49}\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{\dfrac{20}{49}}{3}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=3n_{Al_2O_3}=0,3\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=\dfrac{53}{490}\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=\dfrac{53}{490}.98=10,6\left(g\right)\)

b, Theo PT:  \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)

c, Ta có: m _{dd sau pư} = m_Al2O3 + m _{dd H2SO4} = 210,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{10,6}{210,2}.100\%\approx5,04\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{210,2}.100\%\approx16,3\%\end{matrix}\right.\)

Bạn tham khảo nhé!