BTKL :
\(m_{O_2} = m_B - m_A = 22,9-19,3 = 3,6(gam)\\ \Rightarrow n_{O_2} = \dfrac{3,6}{32} = 0,1125(mol)\)
\(n_{SO_2} = \dfrac{11,76}{22,4} = 0,525(mol)\)
Gọi : \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)⇒ 27a + 56b = 19,3(1)
Bảo toàn e :
\(3n_{Al} + 3n_{Fe} = 4n_{O_2} + 2n_{SO_2}\\ \Rightarrow 3a + 3b = 0,1125.4 + 0,525.2(2)\)
Từ (1)(2) suy ra : a = 0,3 ; b = 0,2
Vậy : \(m_{Fe} = 0,2.56 = 11,2(gam) \)
(Đáp án D)
\(m_{O_2}=22.9-19.3=3.6\left(g\right)\)
\(n_{O_2}=\dfrac{3.6}{32}=0.1125\left(mol\right)\)
\(n_{SO_2}=0.525\left(mol\right)\)
\(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)
\(m=27a+56b=19.3\left(g\right)\left(1\right)\)
\(Al\rightarrow Al^{3+}+3e\)
\(Fe\rightarrow Fe^{3+}+3e\)
\(O_2+4e\rightarrow2O^{2-}\)
\(S^{+6}+2e\rightarrow S^{+4}\)
\(BTe:\)
\(3a+3b=0.1125\cdot4+0.525\cdot2=1.5\left(2\right)\)
\(\left(1\right),\left(2\right):\)
\(a=0.3,b=0.2\)
\(m_{Fe}=0.2\cdot56=11.2\left(g\right)\Rightarrow D\)
\(n_{O_2}=0,1125\left(mol\right)\)
\(Al^0-3e\rightarrow Al^{+3}\)
.x.......3x............
\(Fe^0-3e\rightarrow Fe^{+3}\)
.y.......3y.........
\(S^{+6}+2e\rightarrow S^{+4}\)
.0,525..1,05...
\(O^0_2+4e\rightarrow2O^{-2}\)
.0,1125..0,45.....
Bte : \(3x+3y=1,5\)
PT KL : \(27x+56y=19,3\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\) ( mol )
=> mFe = 11,2 ( g )
=> Đáp án D .
