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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: \(\Rightarrow\left\{{}\begin{matrix}3x+y=0,2\\27x+56y=5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{470}\\y=\dfrac{37}{470}\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{5,5}\cdot100\%=19,84\%\)
\(\%m_{Fe}=100\%-19,84\%=80,16\%\)
2Al + 3H2SO4 →Al2(SO4)3 + 3H2 (1)
Zn + H2SO4 →ZnSO4 + H2 (2)
a;nH2=\(\dfrac{8,96}{22,4}\)=0,4(mol)
Đặt nAl=a
nZn=b
Ta có:
\(\left\{{}\begin{matrix}27x+65y=11,9\\1,5x+y=0,4\end{matrix}\right.\)
=>a=0,2;b=0,1
mAl=27.0,2=5,4(g)
%mAl=\(\dfrac{5,4}{11,9}.100\)=54,4%
%mZn=54,6%
Ta có kim loại + H2SO4 → muối + H2
nH2 = 0,4 mol
Bảo toàn nguyên tố H có nH2 = nH2SO4 = 0,4 mol
Bảo toàn khối lượng có mkim loại + mH2SO4 = mH2 + mmuối → 11,9 + 0,4.98 = 0,4.2 + m → m = 50,3
a, Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
a---->1,5a--------------------------->1,5a
Mg + H2SO4 ---> MgSO4 + H2
b------>b----------------------->b
Hệ pt \(\left\{{}\begin{matrix}27a+24b=6,3\\1,5a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,15\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,1.27=2,7\left(g\right)\\m_{Mg}=0,15.24=3,6\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{2,7}{6,3}=42,86\%\\\%m_{Mg}=100\%-42,86\%=57,14\%\end{matrix}\right.\)
b, \(n_{H_2SO_4}=0,1.1,5+0,15=0,3\left(mol\right)\)
\(\rightarrow V_{ddH_2SO_4}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)
c, đề yêu cầu jv?
1. Gọi nAl = a (mol)
=> nFe = 1,5a (mol)
PTHH:
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
a ---> 1,5a ---> a ---> 1,5a
Fe + H2SO4 -> FeSO4 + H2
1,5a ---> 1,5a ---> 1,5a ---> 1,5a
=> 342a + 152 . 1,5a = 39,9
=> a = 0,07 (mol)
mAl = 0,07 . 27 = 1,89 (g)
mFe = 0,07 . 1,5 . 56 = 5,88 (g)
2. nH2 = 1,5 . 0,07 + 1,5 . 0,07 = 0,21 (mol)
nO2 = 0,21 . 2 = 0,42 (mol)
nH2O = 2,7/18 = 0,15 (mol)
PTHH: 2H2 + O2 -> (t°) 2H2O
Mol: 0,15 <--- 0,075 <--- 0,15
VE = (0,21 - 0,15 + 0,42 - 0,075) . 22,4 = 9,072 (l)
mE = (0,42 - 0,075) . 32 + (0,21 - 0,15) . 2 = 11,14 (g)
nE = 0,42 - 0,075 + 0,21 - 0,15 = 0,405 (mol)
M(E) = 11,14/0,405 = 27,5 (g/mol)
d(E/N2) = 27,5/28 = 0,98
Gọi kim loại cần tìm là A
Công thức oxit là A2O
Đặt \(\left\{{}\begin{matrix}n_A=x\left(mol\right)\\n_{A_2O}=y\left(mol\right)\end{matrix}\right.\)
=> \(x.M_A+y\left(2.M_A+16\right)=25,8\)
=> \(x.M_A+2y.M_A+16y=25,8\) (1)
PTHH: 2A + 2H2O --> 2AOH + H2
A2O + H2O --> 2AOH
=> \(\left(x+2y\right)\left(M_A+17\right)=33,6\)
=> \(x.M_A+2y.M_A+17x+34y=33,6\) (2)
(2) - (1) = 17x + 18y = 7,8
=> \(x=\dfrac{7,8-18y}{17}\)
Do x > 0 => \(\dfrac{7,8-18y}{17}>0\Rightarrow0< y< \dfrac{13}{30}\) (3)
Thay vào (1) => 7,8.MA + 16y.MA + 272y = 25,8
=> \(M_A=\dfrac{571,2}{7,8+16y}-17\) (4)
(3)(4) => 21,77 < MA < 56,23
=> \(A\left[{}\begin{matrix}Natri\left(Na\right)\\Kali\left(K\right)\end{matrix}\right.\)
- Nếu A là Na:
=> 23x + 62y = 25,8
Và (x + 2y).40 = 33,6
=> x = 0,03; y = 0,405
\(\left\{{}\begin{matrix}m_{Na}=0,03.23=0,69\left(g\right)\\m_{Na_2O}=0,405.62=25,11\left(g\right)\end{matrix}\right.\)
- Nếu A là K
=> 39x + 94y = 25,8
Và (x + 2y).56 = 33,6
=> x = 0,3; y = 0,15
=> \(\left\{{}\begin{matrix}m_K=0,3.39=11,7\left(g\right)\\m_{K_2O}=0,15.94=14,1\left(g\right)\end{matrix}\right.\)
\(a,m_{rắn}=m_{Cu}=2,7\left(g\right)\\ \Rightarrow m_{\left(Zn,Fe\right)}=12-2,7=9,3\left(g\right)\\ n_{H_2}=0,15\left(mol\right),n_{axit}=2.0,2=0,4\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,15}{1}< \dfrac{0,4}{1}\Rightarrow axit.dư\\ \Rightarrow\left\{{}\begin{matrix}65+56b=9,3\\a+b=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{2,7}{12}.100=22,5\%\\ \%m_{Zn}=\dfrac{0,1.65}{12}.100\approx54,167\%\\ \%m_{Fe}=\dfrac{0,05.56}{12}.100\approx23,333\%\)
\(b,ddA:FeCl_2,ZnCl_2,H_2SO_4\left(dư\right)\\ m_{ddH_2SO_4}=200.1,14=228\left(g\right)\\ m_{ddA}=m_{\left(Zn,Fe\right)}+m_{ddH_2SO_4}-m_{H_2}=9,3+228-0,15.2=237\left(g\right)\)
\(C\%_{ddZnCl_2}=\dfrac{136.0,1}{237}.100\approx5,738\%\\ C\%_{ddFeCl_2}=\dfrac{127.0,05}{237}.100\approx2,679\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,15\right).98}{237}.100\approx10,338\%\)
Đã sửa lần cuối lúc 20:45
1. Gọi mol của Mg và Al là x, y mol
=> 24x + 27y = 12,6 (1)
nH2 = 0,6 mol => x + 1,5y = 0,6 (2)
Từ (1) (2) => x = 0,3 ; y = 0,2
=> %Mg = 57,14%
=> %Al = 42,86%
nH2=13,44/22,4=0,6(mol)
Đặt: nMg=a(mol); nAl=b(mol) (a,b>0)
1) PTHH: Mg + H2SO4 -> MgSO4 + H2
a__________a________a_____a(mol)
2Al +3 H2SO4 -> Al2(SO4)3 + 3 H2
b___1,5b______0,5b____1,5b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+27b=12,6\\a+1,5b=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)
=> mMg=0,3.24=7,2(g)
=>%mMg= (7,2/12,6).100=57,143%
=>%mAl=42,857%
2) mMgSO4=120.a=120.0,3=36(g)
mAl2(SO4)3=342.0,5b=342.0,5.0,2= 34,2(g)
mH2SO4= (0,3+0,2.1,5).98=58,8(g)
=>mddH2SO4=58,8: 14,7%=400(g)
=>mddsau= 12,6+400 - 2.0,6= 411,4(g)
=>C%ddAl2(SO4)3= (34,2/411,4).100=8,313%
C%ddMgSO4=(36/411,4).100=8,751%
a, mchất rắn = mCu = 12,8 (g)
=> mhh (Al, Zn) = 28,5 - 12,8 = 16,7 (g)
\(m_{H_2SO_4}=7,84\%.500=39,2\left(g\right)\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)
PTHH:
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2↑
a----->1,5a---------->0,5a-------->1,5a
Zn + H2SO4 ---> ZnSO4 + H2
b---->b------------>b--------->b
mdd (tăng) = mhh (Al, Zn) - mH2 = 27a + 65a - 2.(1,5a - b) = 24a - 63b = 515 - 500 = 15 (g)
=> Hệ pt \(\left\{{}\begin{matrix}27a+65b=15,7\\24a-63b=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\left(TM\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{12,8}{28,5}.100\%=44,9\%\\\%m_{Al}=\dfrac{0,1.27}{28,5}.100\%=18,9\%\\\%m_{Zn}=100\%-44,9\%-18,9\%=36,2\%\end{matrix}\right.\)
b, \(n_{H_2SO_{4\left(dư\right)}}=0,4-0,1.1,5-0,2=0,05\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,05.342}{515}.100\%=3,32\%\\C\%_{ZnSO_4}=\dfrac{0,2.161}{515}.100\%=6,25\%\\C\%_{H_2SO_{4\left(dư\right)}}=\dfrac{0,05.98}{515}.100\%=0,95\%\end{matrix}\right.\)
anh ơi \(24a+63b=15\) mới đúng chứ anh:)
\(m_{dd\left(tăng\right)}=24a+63b\) nữa:)