a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: \(n_{Zn}=n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

Bạn xem lại xem đề cho bao nhiêu gam hỗn hợp nhé, vì m_Zn đã bằng 13 (g) rồi.

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

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

_____0,15<--------------0,15<---0,15

=> m_Fe = 0,15.56 = 8,4 (g)

=> m_Cu = 11,6 - 8,4 = 3,2 (g)

\(\left\{{}\begin{matrix}\%Fe=\dfrac{8,4}{11,6}.100\%=72,414\%\\\%Cu=\dfrac{3,2}{11,6}.100\%=27,586\%\end{matrix}\right.\)

m_FeSO4 = 0,15.152 = 22,8 (g)

Bài 1:

a+b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{Fe}\)

\(\Rightarrow m_{Fe}=0,2\cdot56=11,2\left(g\right)\) \(\Rightarrow m_{Cu}=6,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{11,2}{17,6}\cdot100\%\approx63,64\%\\\%m_{Cu}=36,36\%\end{matrix}\right.\)

c) Ta có: \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

Bảo toàn nguyên tố: \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)=n_{CuSO_4}\)

\(\Rightarrow m_{muối}=0,1\cdot400+0,1\cdot160=56\left(g\right)\)

Bài 2:

Quy đổi hh gồm Fe (a mol) và O (b mol)

\(\Rightarrow56a+16b=27,6\)  (1)

Ta có: \(n_{SO_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

Bảo toàn electron: \(3n_{Fe}=2n_O+2n_{SO_2}\) \(\Rightarrow3a-2b=0,45\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,39\\b=0,36\end{matrix}\right.\)

Bảo toàn nguyên tố: \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,195\left(mol\right)\) \(\Rightarrow m_{Fe_2\left(SO_4\right)_3}=0,195\cdot400=78\left(g\right)\)

Câu 1:

a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(2K+2H_2O\rightarrow2KOH+H_2\)

b, Ta có: \(23n_{Na}+39n_K=4,18\left(1\right)\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{1}{2}n_K=\dfrac{1,568}{22,4}=0,07\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na}=0,08\left(mol\right)\\n_K=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=0,08.23=1,84\left(g\right)\\m_K=0,06.39=2,34\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{NaOH}=n_{Na}=0,08\left(mol\right)\Rightarrow m_{NaOH}=0,08.40=3,2\left(g\right)\)

Câu 2:

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

\(n_{BaO}=\dfrac{15,3}{153}=0,1\left(mol\right)\)

PT: \(Na_2O+H_2O\rightarrow2NaOH\)

______0,2_______________0,4 (mol)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

0,1___________________0,1 (mol)

⇒ m = m_NaOH + m_Ba(OH)2 = 0,4.40 + 0,1.171 = 33,1 (g)

câu 1

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\)

đặt `n_(Na)=a`, `n_K=b`

\(PTHH:2Na+2H_2O->2NaOH+H_2\)

tỉ lệ            2   :       2       :       2       ;       1

n(mol)       a----->a------------>a------------>`1/2a`

\(PTHH:2K+2H_2O->2KOH+H_2\)

Tỉ lệ          2   :     2        :      2        ;   1

n(mol)       b-->b------------->b-------->`1/2b`

ta có

\(\left\{{}\begin{matrix}23a+39b=4,18\\\dfrac{1}{2}a+\dfrac{1}{2}b=0,07\end{matrix}\right.< =>\left\{{}\begin{matrix}a=0,08\\b=0,06\end{matrix}\right.\)

\(m_{Na}=n\cdot M=0,08\cdot23=1,84\left(g\right)\\ m_K=0,06\cdot39=2,34\left(g\right)\\ m_{NaOH}=n\cdot M=0,08\cdot\left(23+16+1\right)=3,2\left(g\right)\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

\(\Rightarrow m_{Ag}=20-13=7\left(g\right)\)

b, \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{20}.100\%=65\%\\\%m_{Ag}=100-65=35\%\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}n_{Fe_2O_3}=x\left(mol\right)\\n_{PbO}=y\left(mol\right)\end{matrix}\right.\)

\(PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ \left(mol\right)......x\rightarrow...3x......2x.....3x\\ PTHH:PbO+H_2\underrightarrow{t^o}Pb+H_2O\\ \left(mol\right)......y\rightarrow.y.....y......y\\ m_{Fe_2O_3}+m_{PbO}=\Sigma m_{hh}\\ \Leftrightarrow160x+223y=76,6\left(1\right)\\ m_{Fe}+m_{Pb}=\Sigma m_{kl}\\ \Leftrightarrow56.2x+207y=63,8\\ \Leftrightarrow112x+207y=63,8\left(2\right)\\ \xrightarrow[\left(1\right)]{\left(2\right)}\left\{{}\begin{matrix}160x+223y=76,6\\112x+207y=63,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{0,2.160}{76,6}.100\%=41,8\%\\\%m_{PbO}=100\%-41,8\%=58,2\%\end{matrix}\right.\)

\(\Sigma n_{H_2}=3x+y=3.0,2+0,2=0,8\left(mol\right)\\ \Sigma V_{H_2}=0,8.22,4=17,92\left(l\right)\)

Câu c là H2 chứ bạn

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(Đặt:n_{Zn}=x\left(mol\right);n_{Fe}=y\left(mol\right)\)

\(n_{H_2}=0,4\left(mol\right)\)

Theo đề ta có hệ \(\left\{{}\begin{matrix}65x+56y=24,2\\x+y=0,4\end{matrix}\right.\)

=> x=0,2 ; y=0,2

\(\%m_{Zn}=\dfrac{0,2.65}{24,2}.100=53,72\%;\%m_{Fe}=46,28\%\)

b)Bảo toàn nguyên tố H: \(n_{HCl}=2n_{H_2}=0,8\left(mol\right)\)

=> \(V_{HCl}=\dfrac{0,8}{2,5}=0,32\left(l\right)\)

c) \(n_{FeCl_2}=0,2\left(mol\right);n_{ZnCl_2}=0,2\left(mol\right)\)

=> \(CM_{FeCl_2}=\dfrac{0,2}{0,32}=0,625\left(mol\right)\)

\(CM_{ZnCl_2}=\dfrac{0,2}{0,32}=0,625\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

⇒ m_Zn = 0,2.65 = 13 (g)

⇒ m_Cu = 19,4 - 13 = 6,4 (g)

Bạn tham khảo nhé!