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\(n_{FeSO_4}=\dfrac{m}{M}=\dfrac{22,8}{152}=0,15\left(mol\right)\\ PTHH:Fe+H_2SO_4->FeSO_4+H_2\)
tỉ lệ 1 : 1 : 1 : 1
n(mol) 0,15<-----------------0,15
\(m_{Fe}=n\cdot M=0,15\cdot56=8,4\left(g\right)\\ =>B\)
Cu(OH)2 + H2SO4 \(\rightarrow\) CuSO4 + 2H2O
nCu(OH)2 = \(\dfrac{29,4}{98}=0,3mol\)
Theo pt: nH2SO4 = nCu(OH)2 = 0,3 mol
=> mH2SO4 = 0,3.98 = 29,4g
VH2SO4 = 0,3:1 = 0,3l
\(n_{Mg}=\dfrac{10,8}{24}=0,45\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,45 0,45 0,45
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\\
C\%_{H_2SO_4}=\dfrac{44,1}{176,4}.100\%=25\%\\
V_{H_2}=0,45.22,4=10,08\left(l\right)\)
Bài 1:
PTHH: \(Na_2CO_3+BaCl_2\rightarrow2NaCl+BaCO_3\downarrow\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=\dfrac{200\cdot5,3\%}{106}=0,103\left(mol\right)\\n_{BaCl_2}=\dfrac{300\cdot10,4\%}{208}=0,15\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,103}{1}< \dfrac{0,15}{1}\) \(\Rightarrow\) BaCl2 còn dư, Na2CO3 p/ứ hết
\(\Rightarrow n_{BaCO_3}=0,103\left(mol\right)\) \(\Rightarrow m_{BaCO_3}=0,103\cdot197=20,291\left(g\right)\)
b) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=0,206\left(mol\right)\\n_{BaCl_2\left(dư\right)}=0,047\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,206\cdot58,5=12,051\left(g\right)\\m_{BaCl_2\left(dư\right)}=0,047\cdot208=9,776\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddNa_2CO_3}+m_{ddBaCl_2}-m_{BaCO_3}=479,709\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{12,051}{479,709}\cdot100\%\approx2,51\%\\C\%_{BaCl_2\left(dư\right)}=\dfrac{9,776}{479,709}\cdot100\%\approx2,04\%\end{matrix}\right.\)
Bài 2:
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) (1)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\) (2)
a) Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+27b=9,38\) (*)
Ta có: \(n_{H_2}=\dfrac{6,944}{22,4}=0,31\left(mol\right)\) \(\Rightarrow a+\dfrac{3}{2}b=0,31\) (**)
Từ (*) và (**) \(\Rightarrow\left\{{}\begin{matrix}a=n_{Fe}=0,1\left(mol\right)\\b=n_{Al}=0,14\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,1\cdot56=5,6\left(g\right)\\m_{Al}=3,78\left(g\right)\end{matrix}\right.\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Fe}=0,2\left(mol\right)\\n_{HCl\left(2\right)}=3n_{Al}=0,42\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{HCl}=0,62\left(mol\right)\) \(\Rightarrow C\%_{HCl}=\dfrac{0,62\cdot36,5}{200}\cdot100\%=11,315\%\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{FeCl_2}=0,1\left(mol\right)\\n_{AlCl_3}=0,14\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\m_{AlCl_3}=0,14\cdot133,5=18,69\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{H_2}=0,31\cdot2=0,62\left(g\right)\)
\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{H_2}=208,76\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{12,7}{208,76}\cdot100\%\approx6,08\%\\C\%_{AlCl_3}=\dfrac{18,69}{208,76}\cdot100\%\approx8,95\%\end{matrix}\right.\)
2Fe(OH)3+3H2SO4(loãng)→ Fe2(SO4)3+ 6H2O
(mol) 0,15 0,225
\(n_{Fe\left(OH\right)_3}=\dfrac{m}{M}=\dfrac{16,05}{107}=0,15\left(mol\right)\)
\(->m_{H_2SO_4}=n.M=0,225.98=22,05\left(g\right)\)
Ta có:
\(C\%=\dfrac{m_{H_2SO_4}}{m_{ddH_2SO_4}}.100\%=7,35\%\)
<=> \(m_{ddH_2SO_4}=\dfrac{22,05.100}{7,35}=300\left(g\right)\)
Chọn câu: A