\(Fe\left(0,1\right)+CuSO_4\left(0,1\right)\rightarrow FeSO_4\left(0,1\right)+Cu\left(0,1\right)\)

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

\(n_{Fe}=\frac{8,4}{56}=0,15\)

\(m_{ddđ}=100.1,08=108\)

\(n_{CuSO_4}=0,1.1=0,1\)

Ta thấy \(\frac{0,15}{1}>\frac{0,1}{1}\) nên Fe còn dư CuSO₄ hết

\(\Rightarrow a=m_{Cu}=0,1.64=6,4\)

\(\Rightarrow m_{Fe\left(pứ\right)}=0,1.56=5,6\)

\(\Rightarrow m_Y=108+5,6-6,4=107,2\)

\(m_{FeSO_4}=0,1.152=15,2\)

\(\Rightarrow C\%\left(FeSO_4\right)=\frac{15,2}{107,2}.100\%=14,18\%\)

a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(m_{CuCl_2}=270.10\%=27\left(g\right)\Rightarrow n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)

Ta có: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\) ⇒ Fe hết, CuCl₂ dư

PTHH: Fe + CuCl₂ ---> FeCl₂ + Cu

Mol:    0,15    0,15          0,15     0,15

\(a=m_{Cu}=0,15.64=9,6\left(g\right)\)

b, \(m_{dd.sau.pứ}=8,4+270-9,6=268,8\left(g\right)\)

\(m_{CuCl_2dư}=\left(0,2-0,15\right).135=6,75\left(g\right)\)

\(\left\{{}\begin{matrix}C\%_{CuCl_2dư}=\dfrac{6,75.100\%}{268,8}=2,51\%\\C\%_{FeCl_2}=\dfrac{0,15.127.100\%}{268,8}=7,09\%\end{matrix}\right.\)

c, \(V_{ddCuCl_2}=\dfrac{270}{1,35}=200\left(ml\right)=0,2\left(l\right)\)

\(\left\{{}\begin{matrix}C_{M_{CuCl_2dư}}=\dfrac{0,2-0,15}{0,2}=0,25M\\C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75M\end{matrix}\right.\)

\(Fe+CuCl_2\rightarrow Cu+FeCl_2\)

mCuCl2 = \(\dfrac{270.10}{100}=27\left(g\right)\)

\(nCuCl_2=\dfrac{27}{64+71}=0,2\left(mol\right)\)

\(nFe=\dfrac{8,4}{56}=0,15\left(mol\right)\)

CuCl dư 

=> n Cu = nFe = 0, 15(mol)

=> a = mCu = 0,15 . 64 = 9,6(g)

mdd = mFe + mCuCl2 - mCu

        = 8,4  + 270  - 9, 6 =  268,8(g)

mFeCl2 = 0,15 . ( 56 + 71 ) = 19,05 (g)

C%ddA = \(\dfrac{19,05.100}{268,8}=7,09\%\)

c) D = 1,35 . 10  = 13,5 g /ml

V dd A = mdd A / D = 13,725 / 13,5 = 61/60 (l)

CM = \(\dfrac{n}{V}=\dfrac{0,15}{\dfrac{61}{60}}=0,15\) mol / l

\(n_{Fe}=\dfrac{8,4}{56}=0,15mol\)

\(m_{CuCl_2}=\dfrac{270\cdot10\%}{100\%}=27g\Rightarrow n_{CuCl_2}=0,2mol\)

\(Fe+CuCl_2\rightarrow FeCl_2+Cu\)

0,15    0,2          0,15       0,15

\(a=m_{Cu}=0,15\cdot64=9,6g\)

\(m_{FeCl_2}=0,15\cdot127=19,05g\)

\(m_{ddFeCl_2}=8,4+270-0,15\cdot64=268,8g\)

\(C\%=\dfrac{19,05}{268,8}\cdot100\%=7,09\%\)

đây ạ

nFe = 8.4/56=0.15 mol

nCu = 6.4/64=0.1 mol

nAgNO3 = 0.35*2=0.7 mol

Fe + 2AgNO3 --> Fe(NO3)2 + 2Ag

0.15___0.3________0.15_____0.3

Cu + 2AgNO3 --> Cu(NO3)2 + 2Ag

0.1____0.2________0.1_______0.2

nAgNO3( còn lại ) = 0.7 - 0.3 - 0.2 = 0.2 mol

Vì : AgNO3 còn dư nên tiếp tục phản ứng với Fe(NO3)2

Fe(NO3)2 + AgNO3 --> Fe(NO3)3 + Ag

Bđ: 0.15________0.2

Pư: 0.15________0.15_______________0.15

Kt: 0___________0.05_______________0.15

Chất rắn : 0.65 (mol) Ag

mAg = 0.65*108 = 70.2g

em là hs lớp 6 ạ mong senpai tick giúp em

B gồm 3 kim loại là Fe, Cu, Ag

\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,03<----------------0,03

Gọi số mol Cu, Ag là a, b (mol)

=> 64a + 108b = 8,12 - 0,03.56 = 6,44 (g) (1)

\(\left\{{}\begin{matrix}n_{Al^{3+}}=\dfrac{0,81}{27}=0,03\left(mol\right)\\n_{Fe^{2+}}=\dfrac{2,8}{56}-0,03=0,02\left(mol\right)\end{matrix}\right.\)

=> \(n_{NO_3^-}=0,03.3+0,02.2=0,13\left(mol\right)\)

\(\left\{{}\begin{matrix}n_{Cu\left(NO_3\right)_2}=a\left(mol\right)\\n_{AgNO_3}=b\left(mol\right)\end{matrix}\right.\)

=> 2a + b = 0,13 (2)

(1)(2) => a = 0,05 (mol); b = 0,03 (mol)

=> \(C_{M\left(Cu\left(NO_3\right)_2\right)}=\dfrac{0,05}{0,1}=0,5M\)

Đáp án B

 nFe = 0,15 mol ; nCu = 0,1 ; nAg+ = 0,7 mol
Fe + 2Ag+ → Fe2+ + 2Ag (1)
0,15→ 0,3 0,15 0,3
Cu + 2Ag+ → Cu2+ + 2Ag
0,1 → 0,2 0,2
Fe2+ + Ag+ → Fe3+ + Ag (3)
0,15 → 0,15 0,15
Từ (1) ; (2) → m = (0,3 + 0,2 + 0,15).108 = 70,2 gam → Đáp án A