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\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
đặt \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
\(PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)
tỉ lệ 1 : 1 : 1 : 1
n(mol) a---->a------------>a---------->a (1)
\(PTHH:2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)
tỉ lệ 2 : 3 ; 1 : 3
n(mol) b-------->3/2b----->1/2b------------>3/2b (2)
Từ (1) và (2) ta có
\(\left\{{}\begin{matrix}65a+27b=3,79\\a+\dfrac{3}{2}b=0,08\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,02\left(mol\right)\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}m_{Zn}=n\cdot M=0,05\cdot65=3,25\left(g\right)\\m_{Al}=n\cdot M=0,02\cdot27=0,54\left(g\right)\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{3,25\cdot100\%}{3,79}\approx85,75\%\\\%m_{Al}=100\%-85,75\approx14,25\%\end{matrix}\right.\)
với (1) thì
\(n_{H_2SO_4\left(1\right)}=a=0,05\left(mol\right)\)
với (2) thì
\(n_{H_2SO_4\left(2\right)}=\dfrac{3}{2}b=\dfrac{3}{2}\cdot0,02=0,03\left(mol\right)\)
\(=>m_{H_2SO_4}=\left(0,05+0,03\right)\cdot98=7,84\left(g\right)\)
nH2 = \(\dfrac{13,44}{22,4}\)= 0,6 (mol)
Gọi x, y lần lượt là số mol của Mg, Al
Mg + H2SO4 ----> MgSO4 + H2
x x x x (mol)
2Al + 3H2SO4 ----> Al2(SO4)3 + 3H2
y \(\dfrac{3}{2}\)y \(\dfrac{1}{2}\) y \(\dfrac{3}{2}\)y (mol)
a, => x + \(\dfrac{3}{2}\)y = 0,6
24x + 27y = 12,6
=> x = 0,3
y = 0,2
=> mMg = 0,3.24 = 7,2 (g)
=> %Mg = \(\dfrac{7,2.100\%}{12,6}\)= 57,14%
=> %Al = 100 - 57,14 = 42,86%
b, => nH2SO4 = 0,3 + \(\dfrac{3}{2}\).0,2 = 0,6 (mol)
=> mH2SO4 = 0,6.98 = 58,8 (g)
\(a,n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,075<-------------------------0,075
Cu không phản ứng với H2SO4 loãng
b, \(m_{Mg}=0,075.24=1,8\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,8}{8}.100\%=22,5\%\\\%m_{Cu}=100\%-22,5\%=77,5\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,075 0,075
\(Cu+H_2SO_4-x->\)
\(m_{Mg}=0,075.24=1,6\left(g\right)\\ m_{Cu}=8-1,6=6,4\left(g\right)\)
\(\%m_{Cu}=\dfrac{6,4}{8}.100\%=80\%\\
\%m_{Mg}=100-80\%=20\%\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ m_{Fe}=0,1.56=5,6\left(g\right)\\ \%m_{Fe}=\dfrac{5,6}{8}.100=70\%\\ \Rightarrow\%m_{Cu}=100\%-70\%=30\%\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Cu}=20-5,6=14,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\\%m_{Cu}=72\%\end{matrix}\right.\)
a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) Gọi x,y là số mol Al, Fe
\(n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Ta có hệ : \(\left\{{}\begin{matrix}27x+56y=0,83\\\dfrac{3}{2}x+y=0,02\end{matrix}\right.\)
=> \(x=\dfrac{29}{5700};y=\dfrac{47}{3800}\)
\(\%m_{Al}=\dfrac{\dfrac{27}{5700}.27}{0,83}.100=16,55\%\); \(\%m_{Fe}=100-16,55=83,45\%\)
c)Bảo toàn nguyên tố H: \(n_{H_2SO_4}=n_{H_2}=0,02\left(mol\right)\)
=> \(C\%_{H_2SO_4}=\dfrac{0,02.98}{200}.100=0,98\%\)
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)
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$Mg + H_2SO_4 \to MgSO_4 + H_2$
b) Gọi $n_{Al} =a (mol) ; n_{Mg} = b(mol) \Rightarrow 27a + 24b = 11,1(1)$
Theo PTHH : $n_{H_2} = 1,5a + b = \dfrac{11,2}{22,4} = 0,5(2)$
Từ (1)(2) suy ra : a = 0,1 ; b = 0,35
$\%m_{Al} = \dfrac{0,1.27}{11,1}.100\% = 24,3\%$
$\%m_{Mg} = 100\% - 24,3\% = 75,7\%$
c) $n_{Fe_2O_3} = \dfrac{16}{160} = 0,1(mol)$
$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$
Ta thấy :
$n_{Fe_2O_3} : 1 < n_{H_2} : 3$ nên $H_2$ dư
$n_{Fe} = 2n_{Fe_2O_3} = 0,2(mol)$
$m_{Fe} = 0,2.56 = 11,2(gam)$
Mg+H2SO4→→MgSO4+H2
2Al+3H2SO4→→Al2(SO4)3+3H2
nH2=V22,4=13,4422,4=0,6molnH2=V22,4=13,4422,4=0,6mol
- Gọi số mol Mg là x, số mol Al là y. Ta có hệ:
{24x+27y=12,6x+1,5y=0,6⇔{x=0,3y=0,2{24x+27y=12,6x+1,5y=0,6⇔{x=0,3y=0,2
%Al=0,2.27.10012,6≈42,86%%Al=0,2.27.10012,6≈42,86%
%Mg=100%-42,86%=57,14%
nH2SO4=nH2=0,6molnH2SO4=nH2=0,6mol
mH2SO4=0,6.98=58,8g