Đáp án C

Gọi a là số mol H₂ phản ứng

Bảo toàn H: \(n_{H_2O}=n_{H_2}=a\left(mol\right)\)

Bảo toàn khối lượng: \(24+2a=17,6+18a\)

\(\Rightarrow a=0,4\left(mol\right)\)

\(\Rightarrow m_{H_2O}=7,2\left(gam\right)\)

Nguyễn Thị Kiều

Gọi số mol của CuO và Fe₂O₃ là a, b (mol)

=> 80a + 160b = 24

PTHH: CuO + H₂ --t^o--> Cu + H₂O

______a------------------>a---->a

Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

_b-------------------->2b----->3b

=> 64a + 56.2b = 17,6 

=> 64a + 112b = 17,6

=> a = 0,1; b = 0,1

n_H2O = a +3b = 0,4 (mol)

=> m_H2O = 0,4.18 = 7,2(g)

\(m_{CO_2}=17,6g\Rightarrow n_{CO_2}=0,4\left(mol\right)\)

⇒n_{O trong hỗn hợp}=0,4 mol

\(đặtn_{Fe_2O_3}=a;n_{CuO}=b\)

\(\Rightarrow\left\{{}\begin{matrix}160a+80b=24\\3a+b=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,2\\n_{Cu}=0,1\end{matrix}\right.\)

m_O(oxit)=24-17,6=6,4 gam\(\rightarrow\)n_O=\(\dfrac{6,4}{16}=0,4\left(mol\right)\)

CO+O_(oxit)\(\rightarrow\)CO₂

\(n_{CO_2}=n_{O\left(oxit\right)}=0,4\left(mol\right)\rightarrow\)\(m_{CO_2}=0,4.44=17,6\left(gam\right)\)

Fe2O3+3CO--->2Fe+3CO2

x_____________2x

CuO+CO--->Cu+CO2

y___________y

Hệ pt:

\(\left\{{}\begin{matrix}160x+80y=24\\112x+64y=17,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

=>mFe2O3=0,1.160=16(g)

=>%mFe2O3=16/24.100%=66,67%

=>%mCuO=100-66,67=33,33%

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Đặt:n_{Fe}=a\left(mol\right);n_{Cu}=0,5a\left(mol\right)\\ m_{hhB}=17,6\\ \Leftrightarrow56a+64.0,5a=17,6\\ \Leftrightarrow a=0,2\left(mol\right)\\ \Rightarrow n_{Fe}=0,2\left(mol\right);n_{Cu}=0,1\left(mol\right)\\ a,n_{H_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,2+0,1=0,4\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ b,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ \Rightarrow ddC:FeCl_2,HCldư\\ n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_2}=0,2.127=25,4\left(g\right)\\ n_{HCl\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

Chọn C

m o x i t   =   m K L   +   m o x i   →   m o x i   =   m o x i t   –   m K L     =   24   –   17 , 6   =   6 , 4   g a m .