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a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$
b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)
Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$
Bài 6:
\(n_{Fe\left(OH\right)_3}=\dfrac{21,4}{107}=0,2\left(mol\right)\)
PT: \(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)
_______0,2________0,6______0,2 (mol)
a, \(C\%_{HCl}=\dfrac{0,6.36,5}{200}.100\%=10,95\%\)
b, \(C\%_{FeCl_3}=\dfrac{0,2.162,5}{21,4+200}.100\%\approx14,68\%\)
Bài 7:
\(m_{H_2SO_4}=100.9,8\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)
PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
______0,1______0,1_______0,1 (mol)
a, \(m_{ZnO}=0,1.81=8,1\left(g\right)\)
b, \(C\%_{ZnSO_4}=\dfrac{0,1.161}{8,1+100}.100\%\approx14,89\%\)
\(a.n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\2 K+2H_2O\xrightarrow[]{}2KOH+H_2\\ n_K=0,5.2=1\left(mol\right)\\ m_K=1.39=39\left(g\right)\\b.n_{KOH}=n_K=1mol\\ m_{KOH}=1.56=56\left(g\right)\\ m_{H_2}=0,5.2=1\left(g\right)\\ m_{ddKOH}=200+39-1=238\left(g\right)\\ C_{\%KOH}=\dfrac{56}{238}\cdot100=23,53\%\)
nH2=13,14:22,4=0,6 mol
PTHH: 2Al+6HCl=>2Al2Cl3+3H2
0,4<-1,2<----0,4<-----0,6
=> Al=0,4.27=10,8g
CMHCL=1,2:0,4=3M
CM Al2Cl3=0,4:0,4=1M
bài 2: nH2=0,2mol
PTHH: 2A+xH2SO4=> A2(SO4)x+xH2
0,4:x<---------------------------0,2
ta có PT: \(\frac{13}{A}=\frac{0,4}{x}\)<=> 13x=0,4A
=> A=32,5x
ta lập bảng xét
x=1=> A=32,5 loiaj
x=2=> A=65 nhận
x=3=> A=97,5 loại
=> A là kẽm (Zn)
Mg+2CH3COOH->(CH3COO)2Mg+H2
0,15------0,3-------------0,15-------------0,15
n Mg=\(\dfrac{3,6}{24}\)=0,15 mol
m CH3COOH=24g =>n CH3COOH=\(\dfrac{24}{60}\)=0,6 mol
->CH3COOH dư
=>C% (CH3COO)2Mg=\(\dfrac{0,15.142}{200+3,6-0,15.2}\).100=10,48%
=>C% CH3COOH dư= \(\dfrac{0,3.60}{200+3,6-0,15.2}\).100=8,85%
Bài 2:
a) PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
b) Dung dịch A là dung dịch bazơ
Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\) \(\Rightarrow n_{NaOH}=0,1\left(mol\right)\) \(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{1}=0,1\left(M\right)\)
c) Sửa đề: dd H2SO4 9,8%
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PTHH: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,05\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,05\cdot98}{9,8\%}=50\left(g\right)\) \(\Rightarrow V_{ddH_2SO_4}=\dfrac{50}{1,14}\approx43,86\left(ml\right)\)
Bài 1:
PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot19,6\%}{98}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư
\(\Rightarrow n_{CuSO_4}=0,2\left(mol\right)=n_{H_2SO_4\left(dư\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{0,2\cdot160}{200+16}\cdot100\%\approx14,81\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2\cdot98}{200+16}\cdot100\%\approx9,07\%\end{matrix}\right.\)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=\dfrac{200\cdot4\%}{40}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{150\cdot9,8\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) \(\Rightarrow\) Axit còn dư, tính theo Bazơ
\(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_4}=\dfrac{0,1\cdot142}{200+150}\cdot100\%\approx4,06\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{200+150}\cdot100\%=1,4\%\end{matrix}\right.\)
$\rm n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$
Phương trình hóa học
$\rm Fe+ 2HCl \to FeCl_2 +H_2$
Theo phương trình hóa học
$\rm n_{FeCl_2}=n_{H_2}=n_{Fe}=0,2(mol)$
$\rm m_{dung~dịch~sp ứ} =200+11,2-0,2.2=210,8(g)$
`->` $\rm C\%_{FeCl_2}=\dfrac{0,2.127}{210,8}.100\%=12,05\%$