1) Dễ thấy \(V_{H_2\left(P_1\right)}< V_{H_2\left(P_2\right)}\Rightarrow\) Ở phần 1 Al dư
Gọi \(\left\{{}\begin{matrix}n_K=x\left(mol\right)\\n_{Al}=y\left(mol\right)\\n_{Fe}=z\left(mol\right)\end{matrix}\right.\)
`-` Phần 1: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: \(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)
x-------------------x--------->0,5x
\(2KOH+2Al+2H_2O\rightarrow2KAlO_2+3H_2\uparrow\)
x--------------------------------------------->1,5x
`=> 1,5x + 0,5x = 0,2 => x = 0,1 (1)`
`-` Phần 2: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
PTHH: \(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)
x--------------------------->0,5x
\(2KOH+2Al+2H_2O\rightarrow2KAlO_2+3H_2\uparrow\)
y------------------------------->1,5y
`=>` 0,5x + 1,5y = 0,35 (2)`
`-` Phần 3: \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,5.1,2=0,6\left(mol\right)\\n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\end{matrix}\right.\)
PTHH: \(2K+H_2SO_4\rightarrow K_2SO_4+H_2\uparrow\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{H_2}=0,45\left(mol\right)< 0,6\left(mol\right)\)
`=> H_2SO_4` dư, hh kim loại tan hết
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_K+\dfrac{3}{2}n_{Al}+n_{Fe}\Rightarrow0,5x+1,5y+z=0,45\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\\z=0,1\end{matrix}\right.\)
`=>` \(E\left\{{}\begin{matrix}m_K=0,1.39=3,9\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)
2) \(n_{NaOH}=\dfrac{240.20\%}{40}=1,2\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(dư\right)}=0,6-0,45=0,15\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
Giả sử NaOH dư
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,3<--------0,15
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3\downarrow+3Na_2SO_4\)
0,1--------->0,6---------->0,2
\(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+Na_2SO_4\)
0,1------->0,2---------->0,1
`=>` \(n_{NaOH\left(pư\right)}=0,3+0,6+0,2=1,1\left(mol\right)\)
So sánh `1,1 < 1,2 => NaOH` dư, giả sử đúng
`=>` \(n_{NaOH\left(dư\right)}=1,2-1,1=0,1\left(mol\right)\)
Xảy ra phản ứng: \(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+2H_2O\)
ban đầu 0,1 0,2
phản ứng 0,1-------->0,1
sau phản ứng 0 0,1
PTHH: \(2Al\left(OH\right)_3\xrightarrow[]{t^o}Al_2O_3+3H_2O\)
0,1----------->0,05
\(4Fe\left(OH\right)_2+O_2\xrightarrow[]{t^o}2Fe_2O_3+4H_2O\)
0,1-------------------->0,05
`=> m = 0,05.102 + 0,05.160 = 13,1 (g)`
PTHH:
