\(a)\)
\(Fe_2O3 + 3H_2-t^o-> 2Fe + 3H_2O\)(1)
\(CuO + H_2-t^o-> Cu + H_2O\)(2)
Theo đề: mFe2O3: mCuO = 3: 1.
Gọi \(mCuO =a \)\((g)\) \(=> mFe_2O_3 = 3a \)\((g)\)
Ta có:\(mCuO + mFe_2O_3 = 48 (g)\)
\(<=> 3a + a = 48 \)
\(=> a =12 \)
\(=>\) \(mFe_2O_3 \)\(=3a = 36 (g)\)\(=> nFe2O3 = 0,225 (mol)\)
\(mCuO= a = 12 (g)\)\(=> nCuO =0,15 (mol)\)
Vậy \(nH_2 \) đã dùng = \(nH_2 \)(1) + \(nH_2 \)(2)
\(<=>nH2 = 0,675 + 0,15 = 0,825 (mol)\)
Thể tích khí H2 là:
\(V_H2 (đktc) = 22,4.nH2 \) \(= 22,4. 0,825 = 18,48 (l)\)
\(b)\)
Theo (1) \(nFe = 2.nFe_2O_3\) = \(2.0,225 = 0,45 (mol)\)
\(=> mFe = 0,45 . 56 = 25,2 (g)\)
Theo (2) \(nCu = nCuO = 0,15 (mol)\)
\(=> mCu = 0,15.64 = 9,6 (g)\)
%mFe = \(\dfrac{mFe .100}{mFe + mCu}\) = \(\dfrac{25,2.100}{25,2+9,6}\) = 72,41%
=> %mCuO = 100% - 72,41% = 27,59%
