E xem lại đề 

a) PTHH: MgO + 2 HCl-> MgCl2 + H2O (1)

CuO +2 HCl -> CuCl2 + H2O (2)

Gọi x,y lần lượt là số mol của MgO và CuO.

Ta có: mhhA= 8(g)

<=> mMgO + mCuO= 8

<=> 40x+80y=8 (a)

mHCl= 150.7,3%= 10,95(g) => nHCl= 10,95/36,5= 0,3(mol)

=> nHCl(1) + nHCl(2)= 0,3

<=> 2x+2y=0,3(b)

Từ (a) và (b) nên:

\(\left\{{}\begin{matrix}40x+80y=8\\2x+2y=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

=> mMgO= 0,1.40=4(g) ; mCuO= 8-4=4(g)

=>%mMgO= (4/8).100= 50% => %mCuO=50%

b) mddB= mhhA+ mddHCl= 8+150=158(g)

mMgCl2= 95x= 95. 0,1= 9,5(g)

mCuCl2=135y=135.0,05=6,75(g)

C%ddMgCl2=(9,5/158).100= 6,013%

C%ddCuCl2= (6,75/158).100 = 4,272%

c) PTHH: (3) MgO + H2SO4 -> MgSO4 + H2O

(4) CuO + H2SO4 -> CuSO4 + H2O

nH2SO4(3+4)= nH2SO4(3) + nH2SO4(4)= nMgO(3) + nCuO(4)= 0,1+0,0,5= 0,15(mol)

=> VddH2SO4= nH2SO4/CMddH2SO4= 0,15/2= 0,075(l)= 75(ml)

=> mddH2SO4= 1,2.75= 90(g)

\(n_{CuO}=\dfrac{32}{80}=0,4(mol)\\ CuO+2HCl\to CuCl_2+H_2\\ \Rightarrow n_{HCl}=0,8(mol);n_{CuCl_2}=n_{H_2}=0,4(mol)\\ a,m_{dd_{HCl}}=\dfrac{0,8.36,5}{20\%}=146(g)\\ b,m_{CuCl_2}=0,4.135=54(g)\\ c,C\%_{CuCl_2}=\dfrac{54}{32+146-0,4.2}.100\%=30,47\%\)

\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)

\(n_{CuO}= \dfrac{32}{80}= 0,4 mol\)

Theo PTHH:

\(n_{HCl}= 2n_{CuO}= 0,8 mol\)

\(\Rightarrow m_{HCl}= 0,8 . 36,5=29,2 g\)

\(\rightarrow m_{dd HCl}= \dfrac{29,2 . 100%}{20%}= 146 g\)

b) Muối tạo thành là CuCl₂

Theo PTHH:

\(n_{CuCl_2}= n_{CuO}= 0,4 mol\)

\(\Rightarrow m_{CuCl_2}= 0,4 . 135= 54g\)

c)

\(m_{dd sau pư}= m_{CuO} + m_{dd HCl}= 32 + 146=178 g\)

C%= \(\dfrac{54}{178} . 100\)%= 30,337 %

\(-\) \(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)

\(pt:CuO+2HCl\rightarrow CUCl_2+H_2O\)

 \(a,\) 

\(-\) \(n_{HCl}=2.n_{CuO}=2.0,4=0,8\left(mol\right)\)

\(\rightarrow\) \(m_{HCl}=0,8.36,5=29,2\left(gam\right)\)

\(\Rightarrow\) \(m_{ddHCl}=29,2:20\%=146\left(gam\right)\)

\(b,\)

\(-\) \(n_{CuCl_2}=n_{CuO}=0,4\left(mol\right)\)

\(\rightarrow\) \(m_{CuCl_2}=0,4.135=54\left(gam\right)\)

\(c,\)

\(m_{ddsaupu}=32+146=178\left(gam\right)\)

\(\Rightarrow\) \(C\%_{CuCl_2}=\dfrac{54}{178}.100\%=30,34\%\)

\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)

\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)