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a. \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(n_{Fe_2O_3}=\frac{48}{160}=0,3mol\)
Từ phương trình \(n_{HCl}=1,8mol\)
\(\rightarrow m_{HCl}=1,8.36,5=65,7g\)
b. \(m_{dd}=m_{Fe_2O_3}+m_{HCl}=48+65,7=113,7g\)
c. \(n_{FeCl_3}=0,6mol\)
\(m_{FeCl_3}=0,6.162,5=97,5g\)
\(C\%_{FeCl_3}=\frac{97,5}{113,7}.100\%\approx85,75\%\)
\(B:ddNaOH\)
\(C:Mg\left(OH\right)_2\)
\(Mg\left(OH\right)_2\rightarrow^{t^o}MgO+H_2O\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
\(n_{Mg\left(OH\right)_2}=n_{Mg}=\frac{6}{40}=0,15mol\)
\(n_{HCl}=\frac{50}{1000}.0,4=0,02mol\)
\(n_{NaOH}=n_{HCl}=0,02mol\)
\(a=0,02.40+0,15.58=9,5g\)
\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)
\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)
nAl= 0,5(mol)
a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2
nHCl= 6/2 . 0,5= 1,5(mol)
=>mHCl= 1,5.36,5=54,75(mol)
=> mddHCl= (54,75.100)/18,25=300(g)
b) nH2= 3/2. 0,5=0,75(mol)
=>V(H2,đktc)=0,75.22,4=16,8(l)
c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)
mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)
=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)
giúp em vs ạ
2HCl+ Na2CO3---->2NaCl+CO2+H2O
n\(_{HCl}=0,1.1=0,1\left(mol\right)\)
m\(_{Na2CO3}=\frac{100.15,9}{100}=15,9\left(g\right)\)
n\(_{Na2CO3}=\frac{15,9}{106}=0,15\left(mol\right)\)
=>Na2CO3 dư
Theo pthh
n\(_{CO2}=\frac{1}{2}n_{HCl}=0,05\left(mol\right)\)
V=V\(_{CO2}=0,05.22,4=1,12\left(l\right)\)
Cho dd Y vào Ba(NO3)2 có kết tủa suy ra Ba(NO3)2 tác dụng vs Na2CO3 dư
Na2CO3+Ba(NO3)2---->BaCO3+2NaNO3
Theo pthh1
n\(_{Na2CO3}=\frac{1}{2}n_{HCl}=0,05\left(mol\right)\)
=> n Na2CO3 dư hay n Na2CO3 tham gia ở pứng 2 là
0,15-0,05=0,1(mol)
Theo pthh2
n\(_{BaCO3}=n_{Na2CO3}=0,1\left(mol\right)\)
m =m\(_{BaCO3}=0,1.197=19,7\left(g\right)\)
\(m_{H_2SO_4}=100.7,84\%=7,84g\)
\(n_{H_2SO_4}=\frac{7,84}{98}=0,08mol\)
BTKL
\(m_Y=m_{Oxit}+m_{H_2SO_4}=2,04+100=102,04g\)
\(m_{H_2SO_4\left(\text{dư}\right)}=102,04.1,92\%=1,959168g\)
\(n_{H_2SO_4\left(\text{dư}\right)}=\frac{1,959168}{98}=0,02mol\)
\(\rightarrow n_{H_2SO_4\text{phản ứng}}=0,08-0,02=0,06mol\)
Đặt Oxit là \(R_2O_n\)
PTHH: \(R_2O_n+n_{H_2SO_4}\rightarrow R_2\left(SO_4\right)_n+n_{H_2O}\)
Theo PTHH \(n_{R_2O_n}=\frac{1}{n}n_{H_2SO_4}=\frac{0,06}{n}\)
\(\rightarrow m_{R_2O_n}=\frac{0,06.\left(2.M_R+n.O\right)}{n}=2,04g\)
\(\rightarrow M_R=\frac{1,08n}{0,12}=9n\)
Biện luận
Với \(n=3\rightarrow M_R=27\)
Vậy Oxit là \(Al_2O_3\)
a. \(Al+2HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
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