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\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Gọi a,b lần lượt là sm của Mg, MgO
Ta có: \(\left\{{}\begin{matrix}24a+40b=6\\a=0,1\end{matrix}\right.\Rightarrow b=0,09\)
\(\Rightarrow\%m_{MgO}=\dfrac{0,09.40}{6}.100=60\%\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\) (1)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (2)
a) Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=n_{Mg}\) \(\Rightarrow m_{Mg}=1\cdot24=24\left(g\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{24}{32}\cdot100\%=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)
b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\dfrac{32-24}{40}=0,4mol\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{HCl}=2,4mol\) \(\Rightarrow m_{ddHCl}=\dfrac{2,4\cdot36,5}{7,3\%}=1200\left(g\right)\)
c) Theo PTHH: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=1,2mol\)
\(\Rightarrow\Sigma m_{MgCl_2}=1,2\cdot95=114\left(g\right)\)
Mặt khác: \(m_{H_2}=1\cdot2=2\left(g\right)\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)
Câu 5 :
\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,1 0,2 0,1 0,1
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,15 0,3 0,15
a) \(n_{Mg}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(m_{Mg}=0,1.24=2,4\left(g\right)\)
\(m_{MgO}=8,4-2,4=6\left(g\right)\)
0/0Mg = \(\dfrac{2,4.100}{8,4}=28,57\)0/0
0/0MgO = \(\dfrac{6.100}{8,4}=71,43\)0/0
b) Có : \(m_{MgO}=6\left(g\right)\)
\(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,2+0,3=0,5\left(mol\right)\)
\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)
\(n_{MgCl2\left(tổng\right)}=0,1+0,15=0,25\left(mol\right)\)
⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)
\(m_{ddspu}=8,4+500-\left(0,1.2\right)=508,2\left(g\right)\)
\(C_{MgCl2}=\dfrac{14,25.100}{508,2}=2,8\)0/0
Chúc bạn học tốt
\(n_{Al}=a\left(mol\right)\)
\(n_{Fe}=b\left(mol\right)\)
\(m=27a+56b=19.3\left(g\right)\left(1\right)\)
\(n_{H^+}=0.2\cdot2+0.2\cdot2.25\cdot2=1.3\left(mol\right)\)
\(2Al+6H^+\rightarrow2Al^{3+}+3H_2\)
\(Fe+2H^+\rightarrow Fe^{2+}+H_2\)
\(n_{H^+}=3a+2b=1.3\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.3,b=0.2\)
\(\%Al=\dfrac{0.3\cdot27}{19.3}\cdot100\%=41.96\%\)
\(\%Fe=58.04\%\)
\(b.\)
\(n_{H_2}=\dfrac{1}{2}n_{H^+}=0.65\left(mol\right)\)
Bảo toàn khối lượng :
\(m_{Muối}=19.3+0.4\cdot36.5+0.45\cdot98-0.65\cdot2=76.7\left(g\right)\)
\(a.Mg+2HCl->MgCl_2+H_2\\ MgO+2HCl->MgCl_2+H_2O\\ b.n_{H_2}=\dfrac{2,24}{22,4}=n_{Mg}=0,1mol\\ \%m_{Mg}=\dfrac{0,1.24}{6}=40\%;\%m_{MgO}=60\%\\ n_{MgO}=\dfrac{0,6.6}{40}=0,09\left(mol\right)\\ n_{MgCl_2}=0,1+0,09=0,19\left(mol\right)\\ n_{HCl}=0,19.2=0,38\left(mol\right)\\ V_{ddHCl}=\dfrac{0,38.36,5}{0,2.1,1}=63,0\left(mL\right)\\ C_{M\left(MgCl_2\right)}=\dfrac{0,19}{0,063}=3,0\left(M\right)\)
Gọi số mol của Al và Fe lần lượt là a và b
Ta có 1,5a + b = 0,25
1,5a = 0,15
=> a = 0,1 và b = 0,1
=> %Al = 32,53%
%Fe= 67,47%
nH2=2.24/22.4=0.1 mol
pt: Mg+2HCL->MgCl2+H2
nMg=nH2=0.1 mol
mMg=0.1*24=2.4g
=>mMg=6-2.4=3.6g
%mMg=(3.6/6)*100=60%
chỗ mMg=0,1*24:22,4=2,57g