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\(n_{HCl}=1\cdot0,2=0,2\left(mol\right)\\ PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\\ a,n_{MgO}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\\ \Rightarrow m=m_{MgO}=0,1\cdot40=4\left(g\right)\\ b,n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\ c,m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{250}\cdot100\%=2,92\%\)
\(n_{H_2O}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{H_2O}=0,1\cdot18=1,8\left(g\right)\\ \Rightarrow m_{dd_{MgCl_2}}=4+250-1,8=252,2\left(g\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{252,2}\cdot100\%\approx3,77\%\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a) Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=n_{Mg}\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,5\cdot24}{16}=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Mg}=0,5\left(mol\right)\\n_{MgO}=\dfrac{16\cdot25\%}{40}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=2n_{Mg}+2n_{MgO}=1,2\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{1,2\cdot36,5}{20\%}=219\left(g\right)\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,5\left(mol\right)\\n_{MgCl_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2}=0,5\cdot2=1\left(g\right)\\m_{MgCl_2}=0,6\cdot95=57\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{hhA}+m_{ddHCl}-m_{H_2}=234\left(g\right)\) \(\Rightarrow C\%_{MgCl_2}=\dfrac{57}{234}\cdot100\%\approx24,36\%\)
Cho mình hỏi ở cái PTHH ấy! sao ta không tính số mol ở dưới??
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo pt: \(\Rightarrow\left\{{}\begin{matrix}3x+y=0,2\\27x+56y=5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{470}\\y=\dfrac{37}{470}\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{5,5}\cdot100\%=19,84\%\)
\(\%m_{Fe}=100\%-19,84\%=80,16\%\)
Answer:
\(PTHH:\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(x\) \(2x\) \(x\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(y\) \(2y\) \(y\)
Gọi x và y lần lượt là số mol của \(MgO,CuO\)
\(nHCl=\frac{14,6}{36,5}=0,4mol\)
a.
\(\left(1\right)40x+80y=12\)
\(\left(2\right)2x+2y=0,4\)
Giải hệ phương trình \(\left(1\right)\) và \(\left(2\right)\Rightarrow x=y=0,1\)
\(mMgO=0,1.40=4g\)
\(mCuO=12-4=8g\)
b.
\(mMgCl_2=0,1.95=9,5g\)
\(mCuCl_2=0,1.135=13,5g\)
c.
\(mddsaupu=mhh+mddHCl=12+14,6=26,6g\)
\(C\%_{MgCl_2}=\frac{9,5.100}{26,6}=35,71\%\)
\(C\%_{CuCl_2}=\frac{13,5.100}{26,6}=50,75\%\)