PT: \(CuO+C\underrightarrow{t^o}Cu+CO\)
\(Fe_2O_3+3C\underrightarrow{t^o}2Fe+3CO\)
\(2Al_2O_3+9C\underrightarrow{t^o}Al_4C_3+6CO\)
\(CaO+3C\underrightarrow{t^o}CaC_2+CO\)
\(Al_4C_3+12H_2O\rightarrow4Al\left(OH\right)_3+3CH_4\)
\(CaC_2+2H_2O\rightarrow Ca\left(OH\right)_2+C_2H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)
\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
\(2Cu+O_2\underrightarrow{t^o}CuO\)
Ta có: \(n_{CuO}=n_{Cu}=n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
⇒ mCuO = 0,2.80 = 16 (g)
\(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\)
⇒ mFe2O3 = 0,2.160 = 32 (g)
Ta có: \(\left\{{}\begin{matrix}n_{CH_4}+n_{C_2H_2}=\dfrac{9,916}{24,79}=0,4\\16n_{CH_4}+26n_{C_2H_2}=4,625.4.0,4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,3\left(mol\right)\\n_{C_2H_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2O_3}=2n_{Al_4C_3}=2.\dfrac{1}{3}n_{CH_4}=0,2\left(mol\right)\\n_{CaO}=n_{CaC_2}=n_{C_2H_2}=0,1\left(mol\right)\end{matrix}\right.\)
⇒ mAl2O3 = 0,2.102 = 20,4 (g)
mCaO = 0,1.56 = 5,6 (g)
