a) Ta có: \(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,1}{0,5}=0,2\left(M\right)=\left[Na^+\right]=\left[Cl^-\right]\)

b) Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\) 

\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=0,4\left(M\right)\\\left[OH^-\right]=0,8\left(M\right)\end{matrix}\right.\)

c) Ta có: \(n_{H_2SO_4}=0,025\cdot2=0,05\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,125+0,025}\approx0,33\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,66\left(M\right)\\\left[SO_4^{2-}\right]=0,33\left(M\right)\end{matrix}\right.\)

Khj cho B td H2SO4 ko co chat khj thoat ra chung to Al va Zn da pu het. 
nCu(NO3)2=0,03=>nCu[+2]=0,03. 
nAgNO3=0,01=>nAg+=0,01 
goi x,y la so mol Al,Zn. 
Al>Al[+3]+3e 
Zn>Zn[+2]+2e 
=>ne nhuog=3x+2y 
Cu[+2]+2e>Cu 
Ag+ + 1e>Ag 
=>ne nhan=0,03.2+0,01=0,07 
theo dlbt e=>3x+2y=0,07 
27x+65y=1,57 
=>x=0,01,y=0,02 
=>nAl(NO3)3=0,01 
=>mAl(NO3)3=2,13g 
nZn(NO3)2=nZn[+2]=0,02=>mZn(NO3)2=3,78g 
khoi luog Cu va Ag la=0,03.64+0,01.108=3g 
=>kl dd giam la 3-1,57=1,43 
=>kl dd luc sau la 101,43-1,43=100g 
=>C%Al(NO3)3=2,13/100=2,13% 
C%Zn(NO3)2=3,78% 

a)

Gọi $n_{CuO} = a(mol) ; n_{Mg} = b(mol)$
$CuO + 2HNO_3 \to Cu(NO_3)_2 + H_2O$
$3Mg + 8HNO_3 \to 3Mg(NO_3)_2 + 2NO + 4H_2O$

Theo PTHH : 

$n_{HNO_3} = 2a + \dfrac{8}{3}b = 0,2.3 = 0,6(mol)$
$n_{NO} = \dfrac{2}{3}b = \dfrac{1,12}{22,4} = 0,05(mol)$

Suy ra a = 0,2 ; b = 0,075

$m = 0,2.80 + 0,075.24 = 17,8(gam)$

b)

$C_{M_{Cu(NO_3)_2}} = \dfrac{0,2}{0,2} = 1M$
$C_{M_{Mg(NO_3)_2}} = \dfrac{0,075}{0,2} = 0,375M$

a, \(\left[Ca^{2+}\right]=\dfrac{0,15.0,5}{0,15+0,05}=0,375M\)

\(\left[Na^+\right]=\dfrac{0,05.2}{0,15+0,05}=0,5M\)

\(\left[Cl^-\right]=\dfrac{0,15.2.0,5+0,05.2}{0,15+0,05}=1,25M\)

b, \(\left[Fe^{3+}\right]=\dfrac{\dfrac{2.1,6}{400}}{1,5}\approx0,005M\)

\(\left[K^+\right]=\dfrac{\dfrac{2.6,96}{174}}{1,5}\approx0,053M\)

\(\left[SO_4^{2-}\right]=\dfrac{\dfrac{3.1,6}{400}+\dfrac{6,96}{174}}{1,5}\approx0,035M\)

\(a,m_{rắn}=m_{Cu}=2,7\left(g\right)\\ \Rightarrow m_{\left(Zn,Fe\right)}=12-2,7=9,3\left(g\right)\\ n_{H_2}=0,15\left(mol\right),n_{axit}=2.0,2=0,4\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,15}{1}< \dfrac{0,4}{1}\Rightarrow axit.dư\\ \Rightarrow\left\{{}\begin{matrix}65+56b=9,3\\a+b=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{2,7}{12}.100=22,5\%\\ \%m_{Zn}=\dfrac{0,1.65}{12}.100\approx54,167\%\\ \%m_{Fe}=\dfrac{0,05.56}{12}.100\approx23,333\%\)

\(b,ddA:FeCl_2,ZnCl_2,H_2SO_4\left(dư\right)\\ m_{ddH_2SO_4}=200.1,14=228\left(g\right)\\ m_{ddA}=m_{\left(Zn,Fe\right)}+m_{ddH_2SO_4}-m_{H_2}=9,3+228-0,15.2=237\left(g\right)\)

\(C\%_{ddZnCl_2}=\dfrac{136.0,1}{237}.100\approx5,738\%\\ C\%_{ddFeCl_2}=\dfrac{127.0,05}{237}.100\approx2,679\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,15\right).98}{237}.100\approx10,338\%\)

Đã sửa lần cuối lúc 20:45

yên tâm t đg lm t ko lm cx có ng lm à

Đáp án B

nCu = nCuO = x; nCu(NO₃)₂ = y

Dung dịch X chỉ chứa 1 chất tan (CuSO₄) => Ion NO₃^- đã hết

3Cu + 8H⁺ + 2NO₃^-→ 3Cu²⁺ + 2NO + 4H₂O

3y ← 8y ←  2y

O^2- + 2H⁺→ H₂O

x   → 2x

=> %mCu = 30,968% => Chọn B.