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Tính số mol " NO2 ; HNO3 ; H2
đặt n,Cu,Al,Fe lần lượt là x,y,z
=> ...
lập hệ
tự giải nhé !
\(PTHH:\text{Cu+4HNO3}\rightarrow\text{Cu(NO3)2+2NO2+2H2O}\)
\(\text{Fe+2HCl}\rightarrow\text{FeCl2+H2}\)
\(\text{2Al+6HCl}\rightarrow\text{2AlCl3+3H2}\)
Gọi a, b lần lượt là số mol của Fe và Al
Ta có :
\(\left\{{}\begin{matrix}\text{nNO2=0,1(mol)}\\\text{nH2=0,175(mol)}\end{matrix}\right.\)
=>nCu=0,05(mol)
=>mCu=0,05x64=3,2(g)
\(\Rightarrow\left\{{}\begin{matrix}\text{56a+27b=6,95}\\\text{a+3b/2=0,175}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\text{a=0,1}\\\text{b=0,05}\end{matrix}\right.\)
\(\text{%Cu=3,2/10,15x100=31,53%}\)
\(\text{%Fe=0,1x56/10,15x100=55,17%}\)
\(\text{%Al=13,3%}\)
Cu + 4HNO3 -----> Cu(NO3)2 + 2NO2 + 2H2O
Fe+2HCl---->FeCl2+H2
2Al+6HCl---->2AlCl3+3H2
n\(_{NO2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo pthh1
n\(_{Cu}=\frac{1}{2}n_{NO2}=0,05\left(mol\right)\)
%m Cu=\(\frac{0,05.64}{10,15}.100\%=31,53\%\)
m\(_{Fe}+m_{Al}=10,15-3,2=6,95\left(g\right)\)
n\(_{H2}=\frac{3,92}{22,4}=0,175\left(mol\right)\)
Gọi n\(_{Fe}=x,n_{Al}=y\)
Theo bài ra ta có pt
\(\)\(\left\{{}\begin{matrix}56x+27y=6,95\\x+1,5x=0,175\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
%m\(_{Fe}=\frac{0,1.56}{10,15}.100\%=55,17\text{%}\)
% m Al=100-55,17-31,53=13,3%
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Zn}\) \(\Rightarrow m_{Zn}=0,1\cdot65=6,5\left(g\right)\)
\(\Rightarrow\%m_{Zn}=\dfrac{6,5}{10}\cdot100\%=65\%\) \(\Rightarrow\%m_{Cu}=35\%\)
c) Theo PTHH: \(n_{HCl}=2n_{Zn}=0,2mol\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,2\cdot36,5}{5\%}=146\left(g\right)\)
2Al+6HCl---->2AlCl3+3H2
Al2o3+6HCl--->2AlCl3+3H2O
Cu+HCl--> không p/u
2Cu + O2---->2CuO
ncuO=2,75/80=0.034375(mol)
Cứ 2 mol Cu---à 2 mol CuO
0.034375<------0.034375
mCu=0,034375.64=2,2(g)
--->%mCu=2,2.100/10=22%
nH2=3,36/22,4=0,15(mol)
cứ 2 mol Al----->3 mol H2
0.1<-----0.15
mAl :0,1.27=2.7(g)
--->%mAl=2,7.100/10=27%
---->%mAl2o3=100%-27%-22%=51%
\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)
\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)
\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)
a) 2Al + 6HCl --> 2AlCl3 + 3H2
Fe + 2HCl --> FeCl2 + H2
b) Gọi số mol Al, Fe lần lượt là a,b
=> 27a + 56b = 13,9
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
2Al + 6HCl --> 2AlCl3 + 3H2
a----->3a--------->a------->1,5a______(mol)
Fe + 2HCl --> FeCl2 + H2
b------>2b-------->b----->b__________(mol)
=> 1,5a + b = 0,35
=> \(\left\{{}\begin{matrix}a=0,1=>m_{Al}=0,1.27=2,7\left(g\right)\\b=0,2=>m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)
c) nHCl = 3a + 2b = 0,7 (mol)
=> mHCl = 0,7.36,5 = 25,55(g)
=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)
\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)
\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)
a)
$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2o$
b)
Theo PTHH : $n_{Mg} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$m_{Mg} = 0,2.24 = 4,8(gam)$
$m_{MgO} = m_{hh} - m_{Mg} = 12,8 - 4,8 = 8(gam)$
c)
$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$
$n_{HCl} = 2n_{Mg} + 2n_{MgO} = 0,8(mol)$
$m_{dd\ HCl} = \dfrac{0,8.36,5}{14,6\%} = 200(gam)$
Bài 1:
Ta có: \(n_{H_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\)
Bảo toàn Hidro: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\) \(\Rightarrow m_{HCl\left(p.ứ\right)}=0,4\cdot36,5=14,6\left(g\right)\)
Bảo toàn khối lượng: \(m_{muối}=m_{KL}+m_{HCl\left(p.ứ\right)}-m_{H_2}=24,7\left(g\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\left(1\right)\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\left(2\right)\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT (1): \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=m_{hh}-m_{Mg}=4,4-2,4=2\left(g\right)\)
Bạn tham khảo nhé!
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Mg}\)
\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\) \(\Rightarrow m_{MgO}=2\left(g\right)\)
Gọi x, y ,z là số mol Cu, Fe, Al.
=> 64x + 56y + 27z = 10.15 (1)
Khi cho vào HNO3 đặc nguội thì chỉ có Cu phản ứng
Cu + 4HNO3 --------> Cu(NO3)2 + 2NO2 + 2H2O
x--------------------------------------...
=> 2x = nNO2 = 2.24/22.4 = 0.1 (2)
Khi cho vào HCl thì có Cu không phản ứng
Fe + 2HCl ----------> FeCl2 + H2
y--------------------------------------...
2Al + 6HCl ----------> 2AlCl3 + 3H2
z--------------------------------------...
=> y+3z/2 = nH2 = 3.92/22.4=0.175 (3)
Giải (1), (2), (3) => x=0.05, y=0.1, z=0.05
mCu = 0.05*64=3.2 gam
mFe = 0.1*56=5.6gam
mAl = 0.05*27=1.35 gam
%mCu = 3.2*100/10.15=31.5
%mFe = 5.6*100/10.15=55.2
%mAl = 1.35*100/10.15=13.3.
Source: Yahoo.
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