\(X\ gồm\ CO_2(x\ mol)\ và\ CO(y\ mol)\\ n_X = x + y = \dfrac{5,6}{22,4} = 0,25(mol)\\ m_X = 44x + 28y = 0,25.2.18,8(gam)\\ \Rightarrow x = 0,15 ; y = 0,1\\ n_{CO\ pư} =n_{CO_2} = 0,15(mol)\\ \Rightarrow V = (0,15 + 0,1).22,4 = 5,6(lít)\\ Oxit : R_2O_n\\ R_2O_n + nCO \xrightarrow{t^o} nCO_2 + 2R\\ n_{oxit} = \dfrac{n_{CO_2}}{n} = \dfrac{0,15}{n} (mol)\\ \Rightarrow \dfrac{0,15}{n} (2R + 16n) = 8\)
\(\Rightarrow R=\dfrac{56}{3} n\)
Với n =3 thì R = 56(Fe)
Vậy oxit : Fe2O3
\(n_{CO\left(dư\right)}=a\left(mol\right),n_{CO_2}=b\left(mol\right)\)
\(n_X=a+b=\dfrac{5.6}{22.4}=0.25\left(mol\right)\left(1\right)\)
\(m_X=18.8\cdot2\cdot0.25=9.4\left(g\right)\)
\(\Rightarrow28a+44b=9.4\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.15\)
\(R_2O_n+nCO\underrightarrow{t^0}2R+nCO_2\)
\(\dfrac{0.15}{n}......0.15\)
\(M_{R_2O_n}=\dfrac{8}{\dfrac{0.15}{n}}=\dfrac{160}{3}n\)
\(\Rightarrow2R+16n=\dfrac{160n}{3}\)
\(\Rightarrow R=\dfrac{56}{3}n\)
\(n=3\Rightarrow R=56\)
\(Rlà:Fe\)
\(V_{CO}=\left(0.1+0.15\right)\cdot22.4=5.6\left(l\right)\)
