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a:
Cu không tác dụng với HCl
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(m_{Mg}=0.2\cdot24=4.8\left(g\right)\)
\(\%Mg=\dfrac{4.8}{10}=48\%\)
b: \(m_{MgCl_2}=0.2\left(24+35.5\cdot2\right)=19\left(g\right)\)
\(m_{dd\left(Saupư\right)}=4.8+90-0.2\cdot2=94.4\)
=>\(C\%=\dfrac{19}{94.4}\simeq20,13\%\)
Vì Cu không tác dụng với HCl
\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,1 0,2 0,1
a) \(n_{Fe}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
\(m_{Cu}=12-5,6=6,4\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddHCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
c) 0/0Fe = \(\dfrac{5,6.100}{12}=46,67\)0/0
0/0Cu = \(\dfrac{6,4.100}{12}=53,33\)0/0
Chúc bạn học tốt
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a, Ta có: \(n_{H_2}=0,1\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{4,4}.100\%\approx54,55\%\\\%m_{MgO}\approx45,45\%\end{matrix}\right.\)
b, Ta có: mMgO = mhhA - mMg = 2 (g)
\(\Rightarrow n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{MgO}=0,1\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{0,1}{2}=0,05\left(l\right)=50\left(ml\right)\)
Bạn tham khảo nhé!
PTHH
Mg + 2HCl ----> MgCl2 + H2 (1)
MgO + 2HCl -----> MgCl2 + H2O (2)
a) Theo pt(1) n Mg = n H2 = \(\frac{1,12}{22,4}\) = 0,05 (mol)
==> m Mg = 0,005 . 24=1,2 (g)
%m Mg = \(\frac{1,2}{3,2}\). 100%= 37,5%
%m MgO= 100% - 37,5%= 62,5%
b)m dd sau pư = 3,2 + 246,9 - 0,05 . 2=250 (g)
Theo pt(1)(2) n MgCl2(1) = n Mg = 0,05 mol
n MgCl2 (2) = n MgO=\(\frac{3,2-1,2}{40}\)=0,05(mol)
==> tổng n MgCl2 = 0,1 (mol) ---->m MgCl2 = 9,5 (g)
C%(MgCl2)= \(\frac{9,5}{250}\) .100% = 3,8%
a) $Zn + 2HCl \to ZnCl_2 + H_2$
$ZnO + 2HCl \to ZnCl_2 + H_2O$
b)
Theo PTHH : $n_{Zn} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$m_{Zn} = 0,2.65 = 13(gam)$
$m_{ZnO} = 21,1 - 13 = 8,1(gam)$
c) $n_{ZnO} = 0,1(mol)$
Theo PTHH : $n_{HCl} = 2n_{Zn} + 2n_{ZnO} = 0,6(mol)$
$m_{dd\ HCl} = \dfrac{0,6.36,5}{16,6\%} = 132(gam)$
d) $m_{dd\ sau\ pư} = 21,1 + 132 - 0,2.2 = 152,7(gam)$
$n_{ZnCl_2} = n_{Zn} + n_{ZnO} = 0,3(mol)$
$C\%_{ZnCl_2} = \dfrac{0,3.136}{152,7}.100\% = 26,72\%$
\(a.PTHH:\)
\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)
\(MgO+2HCl--->MgCl_2+H_2O\left(2\right)\)
b. ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT(1): \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgO}=12-0,2.24=7,2\left(g\right)\)
\(\Rightarrow\%_{MgO}=\dfrac{7,2}{12}.100\%=60\%\)
c. Ta có: \(n_{hh}=0,2+\dfrac{7,2}{40}=0,38\left(mol\right)\)
Theo PT(1,2): \(n_{HCl}=2.n_{hh}=2.0,38=0,76\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,76.36,5=27,74\left(g\right)\)
\(\Rightarrow m_{dd_{HCl}}=138,7\left(g\right)\)
\(\Rightarrow V_{dd_{HCl}}=126\left(ml\right)\)
Đặt: \(\left\{{}\begin{matrix}x=n_{Fe}\left(mol\right)\\y=n_{Al}\left(mol\right)\end{matrix}\right.\)
\(\sum m_{hh}=11\left(g\right)\Rightarrow56x+27y=11\left(1\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ \left(mol\right)....x\rightarrow..2x........x......x\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ \left(mol\right)....y\rightarrow..3y.........y......1,5y\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ \Rightarrow x+1,5y=0,4\left(2\right)\)
\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
a) \(\%m_{Fe}=\dfrac{56.0,1}{11}=51\%\)
\(\rightarrow\%m_{Al}=49\%\)
b) \(\sum m_{ctHCl}=\left(2.0,1+3.0,2\right).36,5=29,2\left(g\right)\)
\(m_{ddHCl}=\dfrac{29,2.100\%}{10\%}=292\left(g\right)\)
c) \(m_{H_2\uparrow}=\left(1.0,1+1,5.0,2\right).2=0,8\left(g\right)\)
\(m_{ddsaupu}=m_{hh}+m_{ddHCl}-m_{H_2\uparrow}=11+292-0,8=302,2\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{0,1.127}{302,2}.100=4,2\%\\ C\%_{AlCl_3}=\dfrac{0,2.133,5}{302,2}.100=8,8\%\)
Đặt: {x=nFe(mol)y=nAl(mol){x=nFe(mol)y=nAl(mol)
∑mhh=11(g)⇒56x+27y=11(1)∑mhh=11(g)⇒56x+27y=11(1)
PTHH:Fe+2HCl→FeCl2+H2↑(mol)....x→..2x........x......xPTHH:2Al+6HCl→2AlCl3+3H2↑(mol)....y→..3y.........y......1,5yPTHH:Fe+2HCl→FeCl2+H2↑(mol)....x→..2x........x......xPTHH:2Al+6HCl→2AlCl3+3H2↑(mol)....y→..3y.........y......1,5y
nH2=8,9622,4=0,4(mol)⇒x+1,5y=0,4(2)nH2=8,9622,4=0,4(mol)⇒x+1,5y=0,4(2)
(1)−→(2){x=0,1y=0,2→(2)(1){x=0,1y=0,2
a) %mFe=56.0,111=51%%mFe=56.0,111=51%
→%mAl=49%→%mAl=49%
b) ∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)
mddHCl=29,2.100%10%=292(g)mddHCl=29,2.100%10%=292(g)
c) mH2↑=(1.0,1+1,5.0,2).2=0,8(g)mH2↑=(1.0,1+1,5.0,2).2=0,8(g)
mddsaupu=mhh+mddHCl−mH2↑=11+292−0,8=302,2(g)mddsaupu=mhh+mddHCl−mH2↑=11+292−0,8=302,2(g)
