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PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a_______a_______a_____a (mol)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
2b______3b__________b_____3b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}56a+27\cdot2b=11\\a+3b=0,2\cdot2=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{11}\cdot100\%\approx50,91\%\\\%m_{Al}=49,09\%\end{matrix}\right.\)
Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{FeSO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)
a) nH2SO4=0,4(mol)
Đặt: nFe=x(mol); nAl=y(mol) (x,y>0)
PTHH: Fe + H2SO4 -> FeSO4 + H2
x________x______x______x(mol)
2Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2
y____1,5y_______0,5y_______1,5y(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}56x+27y=11\\x+1,5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=> mFe=0,1.56=5,6(g)
=>%mFe=(5,6/11).100=50,909%
=>%mAl= 49,091%
b) V(H2,đktc)=0,4.22,4=8,96(l)
c) nAl2(SO4)3= 0,5y=0,5.0,2=0,1(mol)
nFeSO4=x=0,1(mol)
Vddsau=VddH2SO4=0,2(l)
=>CMddAl2(SO4)3= 0,1/0,2=0,5(M)
CMddFeSO4=0,1/0,2=0,5(M)
\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)
Fe + H2SO4 → FeSO4 + H2
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Gọi x,y lần lượt là số mol Fe, Al
\(\left\{{}\begin{matrix}56x+27y=11\\x+\dfrac{3}{2}y=0,4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=>\(\%m_{Fe}=\dfrac{0,1.56}{11}.100=50,91\%\)
=> %m Al = 100 - 50,91 =49,09 %
b)Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,4\left(mol\right)\)
=> \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)
c) \(CM_{FeSO_4}=\dfrac{0,1}{0,2}=0,5M\)
\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{0,2}{2}}{0,2}=0,5M\)
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 (1)
2 3 1 3 ( mol)
2x 3x x 3x
Mg + H2SO4 -> MgSO4 +H2 (2)
1 1 1 1
y y y y
Gọi số mol của Al là x
=> nAl = 2x, nH2SO4 = 3x, nAl2(SO4)3= x , nH2= 3x
Gọi số mol của Mg là y
=> nMg = y, nH2SO4 = y, nMgSO4 = y, nH2= y
Ta có:
mhh = mAl + mMg
1,41= 27.2x + 24y
1,41= 54x + 24y
V H2 (1) + V H2 (2) = 1,568 (1568ml = 1,568 l)
22,4 . 3x + 22,4y = 1,568
67,2x + 22.4y = 1,568
Có : - 54x + 24y=1,41
- 67,2x + 22.4y = 1,568
Bấm máy tính => x = 0,015
y = 0,025
mAl = 54x = 54 . 0,015 = 0,81 9(g)
mMg = 1,41 - 0,81 = 0,6 (g)
%mAl = 0,81 . 100 / 1,41 = 57,45%
%mMg = 100% - 57,45% = 42,55%
\(A.Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ B.n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,05 0,05 0,05 0,05
\(\%m_{Mg}=\dfrac{0,05.24}{6,4}\cdot100=18,75\%\\ \%m_{Cu}=100-18,75=81,25\%\\ C.m_{ddH_2SO_4}=\dfrac{0,05.98}{20}\cdot100=24,5g\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Theo Pt : \(n_{Mg}=n_{H2SO4}=n_{MgSO4}=n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
b) \(\%m_{Mg}=\dfrac{0,05.24}{6,4}.100\%=18,75\%\)
\(\%m_{Cu}=100\%-18,75\%=81,25\%\)
c) \(m_{H2SO4}=0,05.98=4,9\left(g\right)\)
\(\Rightarrow m_{ddH2SO4}=\dfrac{4.100\%}{20\%}=20\left(g\right)\)
Chúc bạn học tốt
a. PTHH : Mg + 2HCl ➝ MgCl2 + H2 (1)
b. theo bài : nH2 = 3,36 : 22,4 = 0,15 (mol)
theo (1) nMg = nH2 = 0,15 (mol)
➞ mMg = 0,15 ✖ 24 = 3,6 (g)
➞ %mMg = (3,6 : 5)✖100 = 72%
➞ %mCu = 100% - 72% = 28%
c. theo (1) nHCl = 2nH2 = 2✖0,15 = 0,3 (mol)
mHCl = 0,3✖36,5 = 10,95(g)
➜mddHCl = (10,95✖100):14,6 = 75(g)
d. dung dịch Y : MgCl2
mdd(spư)= 3,6+75-0,3 = 78,3(g)
theo (1) nMgCl2 = nH2 = 0,15(mol)
mMgCl2 = 0,15✖95 = 14,25(g)
C%MgCl2 = (14,25 : 78,3)✖100 = 18,199%
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a_____2a______a____a (mol)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b_____2b_______b____b (mol)
Ta lập hệ phương trình: \(\left\{{}\begin{matrix}56a+24b=10,4\\a+b=\dfrac{6,72}{22,4}=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{10,4}\cdot100\%\approx53,85\%\\\%m_{Mg}=46,15\%\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M_{MgCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{H_2}=0,2(mol)\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,1}=2M\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow m_{Cu}=20-5,6=14,4(g)\\ c,\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\ \%m_{Cu}=100\%-28\%=72\%\)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a. PTHH:
\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)
\(Cu+HCl--\times-->\)
b. Theo PT(1): \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)
\(\Rightarrow\%_{m_{Mg}}=\dfrac{4,8}{11,2}.100\%=42,9\%\)
\(\%_{m_{Cu}}=100\%-42,9\%=57,1\%\)
c. Theo PT(1): \(n_{HCl}=2.n_{H_2}=2.0,2=0,4\left(mol\right)\)
PTHH: \(NaOH+HCl--->NaCl+H_2O\left(2\right)\)
Theo PT(2): \(n_{NaOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)
a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
X là khí Hidro
b) Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow n_{Al}=0,2mol\)
\(\Rightarrow\%m_{Al}=\dfrac{0,2\cdot27}{8,64}\cdot100\%=62,5\%\) \(\Rightarrow\%m_{Cu}=37,5\%\)
c) Theo PTHH: \(n_{HCl}=3n_{Al}=0,6mol\)
\(\Rightarrow V_{HCl}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)