Đổi 2,016 dm3 = 2,016 l

nH2 = 2,016/22,4 = 0,09 (mol)

Gọi nFe2O3 = a (mol); nCuO = b (mol)

160a + 80b = 5,6 (g) (1)

PTHH:

Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O

Mol: a ---> 3a ---> 2a ---> 3a

CuO + H2 -> (t°) Cu + H2O

Mol: b ---> b ---> b ---> b

3a + b = 0,09 (mol) (2)

Từ (1) và (2) => a = 0,02 (mol); b = 0,03 (mol)

mFe2O3 = 0,02 . 160 = 3,2 (g)

mCuO = 0,03 . 80 = 2,4 (g)

mH2O = (0,02 . 3 + 0,03) . 18 = 1,62 (g)

mFe = 2 . 0,02 . 56 = 2,24 (g)

mCu = 0,03 . 64 = 1,92 (g)

Ta có: \(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) \(\Rightarrow m_{H_2}=0,225\cdot2=0,45\left(g\right)\)

Bảo toàn nguyên tố: \(n_{H_2}=n_{H_2O}=0,225\left(mol\right)\) \(\Rightarrow m_{H_2O}=0,225\cdot18=4,05\left(g\right)\)

Bảo toàn khối lượng: \(m_{KL}=m_{oxit\:}+m_{H_2}-m_{H_2O}=10,4\left(g\right)\)

C

\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

PTHH: CuO + H₂ → Cu + H₂O

Mol:      x         x         x

PTHH: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

Mol:       y            3y        2y

Ta có hpt:\(\left\{{}\begin{matrix}80x+160y=14\\x+3y=0,225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,075\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(m_{hh.kim.loại}=m_{Cu}+m_{Fe}=0,075.64+2.0,05.56=10,4\left(g\right)\)

\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

PTHH:

\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Theo 2 pthh trên: \(n_{H_2O}=n_{H_2}=0,225\left(mol\right)\)

\(\rightarrow m_{H_2O}=0,225.18=4,05\left(g\right)\\ \rightarrow m_{H_2}=0,225.2=0,45\left(g\right)\)

Áp dụng ĐLBTKL, ta có:

\(m_{oxit\left(CuO,Fe_2O_3\right)}+m_{H_2}=m_{\text{kim loại}\left(Cu,Fe\right)}+m_{H_2O}\\ \rightarrow m_{\text{kim loại}\left(Cu,Fe\right)}=14+0,45-4,05=10,4\left(g\right)\)

CuO + H₂ \(\underrightarrow{to}\) Cu + H₂O (1)

Fe₂O₃ + 3H₂ \(\underrightarrow{to}\) 2Fe + 3H₂O (2)

Fe + 2HCl → FeCl₂ + H₂ (3)

\(n_{H_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

a) Theo pT3:; \(n_{Fe}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,2\times56=11,2\left(g\right)\)

\(\Rightarrow m_{Cu}=36,8-11,2=25,6\left(g\right)\)

\(\Rightarrow n_{Cu}=\frac{25,6}{64}=0,4\left(mol\right)\)

Theo PT1: \(n_{CuO}=n_{Cu}=0,4\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,4\times80=32\left(g\right)\)

Theo PT2: \(n_{Fe_2O_3}=\frac{1}{2}n_{Fe}=\frac{1}{2}\times0,2=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe_2O_3}=0,1\times160=16\left(g\right)\)

b) Theo PT3: \(n_{HCl}=2n_{Fe}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4\times36,5=14,6\left(g\right)\)

CuO + H2 -to-> Cu + H2O (1)

Fe2O3 + 3H2 -to-> 2Fe + 3H2O (2)

Fe + 2HCl --> FeCl2 + H2 (3)

nH2= 4.48/22.4= 0.2 mol

Từ (3) :

=> nFe= 0.2 mol , nHCl = 0.4 mol

mFe= 0.2*56=11.2g

mCu= 36.8-11.2=25.6g

nCu= 25.6/64=0.4 mol

Từ (2) => nFe2O3= 0.1 mol

Từ (1) => nCuO = 0.4 mol

mhh = 0.1*160 + 0.4*80 = 48g

mHCl= 0.4*36.5=14.6g

Em bổ sung khối lượng hỗn hợp ban đầu nhé !

Thiếu đề rồi em 

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Đặt:n_{Fe_2O_3}=a\left(mol\right);n_{CuO}=b\left(mol\right)\left(a,b>0\right)\\ m_{hhoxit}=k\left(g\right)\\ \Rightarrow\left(1\right)160a+80b=k\\ \left(2\right)112a+64b=0,72k\\ \Rightarrow6,4a=12,8b\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{12,8}{6,4}=\dfrac{2}{1}\\ \Rightarrow\%m_{Fe_2O_3}=\dfrac{160.2}{160.2+80.1}.100=80\%\\ \Rightarrow\%m_{CuO}=20\%\)