Question

Answer 1

\(Câu20:\)

\(TN1:\)

\(Đặt:n_{Cu}=x\left(mol\right),n_{Zn}=y\left(mol\right)\)

\(m_{hh}=135x+136y=81.36\left(g\right)\left(.\right)\)

\(TN2:\)

\(Đặt:n_{Cu}=kx\left(mol\right),n_{Zn}=ky\left(mol\right),n_{Al}=kz\left(mol\right)\)

\(n_X=kx+ky=0.5\left(mol\right)\left(1\right)\)

\(\dfrac{n_{Zn}}{n_{Al}}=\dfrac{3}{2}\Rightarrow\dfrac{ky}{kz}=\dfrac{3}{2}\Rightarrow2ky-3kz=0\left(2\right)\)

\(n_{H_2}=ky+1.5kz=\dfrac{13.44}{22.4}=0.6\left(mol\right)\left(3\right)\)

\(\left(1\right),\left(2\right),\left(3\right):\)\(kx=0.2,ky=0.3,kz=0.2\)

\(Từ\left(.\right):\)

\(\Rightarrow135\left(x+y\right)+y=81.36\)

\(\Leftrightarrow135\cdot\dfrac{0.5}{k}+\dfrac{0.3}{k}=81.36\)

\(\Leftrightarrow k=\dfrac{135\cdot0.5+0.3}{81.36}=\dfrac{5}{6}\)

\(KĐ:\)

\(x=0.24,y=0.36\)

\(m_{Cu}=0.24\cdot64=15.36\left(g\right)\)

=> B 

Giải hơi dài nhưng em xem thử nhé !!

Answer 2

\(n_{Cu} = a(mol;n_{Zn} = b(mol)\\ Cu + Cl_2 \xrightarrow{t^o} CuCl_2\\ Zn + Cl_2 \xrightarrow{t^o} ZnCl_2\\ \Rightarrow 135a + 136b = 81,36(1)\)

Trong 0,5 mol X có chứa ak mol Cu và bk mol Zn (k>0)

Suy ra:

 \(n_{Al} = \dfrac{2}{3}n_{Zn} = \dfrac{2}{3}bk(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = bk + \dfrac{2}{3}bk. \dfrac{3}{2} = \dfrac{13,44}{22,4} = 0,6(mol)\\ \Rightarrow bk = 0,3\\ \Rightarrow ak = 0,5 - 0,3 = 0,2\\ \Rightarrow \dfrac{a}{b} = \dfrac{ak}{bk} = \dfrac{0,2}{0,3}= \dfrac{2}{3}(2)\)

Từ (1)(2) suy ra: a = 0,24 ; b =0,36

Vậy : \(m_{Cu} = 0,24.64 = 15,36(gam)\)

Đáp án D

Answer 3

Câu20:Câu20:

Đặt:nCu=x(mol),nZn=y(mol)Đặt:nCu=x(mol),nZn=y(mol)

mhh=135x+136y=81.36(g)(.)mhh=135x+136y=81.36(g)(.)

Đặt:nCu=kx(mol),nZn=ky(mol),nAl=kz(mol)Đặt:nCu=kx(mol),nZn=ky(mol),nAl=kz(mol)

nH2=ky+1.5kz=13.4422.4=0.6(mol)(3)nH2=ky+1.5kz=13.4422.4=0.6(mol)(3)

Từ(.):Từ(.):

⇔k=135⋅0.5+0.381.36=56⇔k=135⋅0.5+0.381.36=56

KĐ:KĐ:

mCu=0.24⋅64=15.36(g)mCu=0.24⋅64=15.36(g)

=> B