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\(a,Đặt:n_{Mg}=g\left(mol\right);n_{Fe}=j\left(mol\right)\left(g,j>0\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24g=56j=9,2\\22,4g+22,4j=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}g=0,15\\j=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\approx39,13\%\\ \Rightarrow\%m_{Fe}\approx60,87\%\\ b,n_{HCl}=2n_{H_2}=\dfrac{2.5,6}{22,4}=0,5\left(mol\right)\\ \Rightarrow C_{MddHCl}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\ c,m_{muối}=m_{FeCl_2}+m_{MgCl_2}=95g+127j=95.0,15+127.0,1=26,95\left(g\right)\)
Fe+2HCl->FeCl2+H2
x-----------------------x mol
Mg+2HCl->MgCl2+H2
y-------------------------y mol
ta có\(\left\{{}\begin{matrix}56x+24y=9,2\\x+y=0,25\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
=>%mFe=\(\dfrac{0,1.56}{9,2}.100\)=60,87%
=>%m Mg=39,13%
Ta có : n HCl=0,1.2+0,15.2=0,5 mol
=>CMHCl=\(\dfrac{0,5}{0,2}\)=2,5M
=>m muối =0,1.127+0,15.95=26,95g
a)
2Al + 6HCl --> 2AlCl3 + 3H2
Fe + 2HCl --> FeCl2 + H2
b)Gọi số mol Al, Fe là a, b (mol)
=> 27a + 56b = 11 (1)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a--->3a-------->a------>1,5a
Fe + 2HCl --> FeCl2 + H2
b---->2b------>b---->b
=> \(n_{H_2}=1,5a+b=\dfrac{8,96}{22,4}=0,4\left(mol\right)\) (2)
(1)(2) => a = 0,2; b = 0,1
mHCl = (0,6 + 0,2).36,5 = 29,2 (g)
=> \(m_{ddHCl}=\dfrac{29,2.100}{9,125}=320\left(g\right)\)
mdd sau pư = 11 + 320 - 0,4.2 = 330,2 (g)
\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{330,2}.100\%=8,086\%\\C\%_{FeCl_2}=\dfrac{0,1.127}{330,2}.100\%=3,846\%\end{matrix}\right.\)
Đặt: nAl = a (mol); nFe = b (mol)
27a + 56b = 11 (g) (1)
nH2 = 8,96/22,4 = 0,4 (mol)
PTHH:
2Al + 6HCl -> 2AlCl3 + 3H2
Mol: a ---> 3a ---> a ---> 1,5a
Fe + 2HCl -> FeCl2 + H2
Mol: b ---> 2b ---> b ---> b
nH2 = 1,5a + b = 0,4 (mol)
Từ (1)(2) <=> a = 0,2 (mol); b = 0,1 (mol)
Còn C% bạn tự tính
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: \(n_{H_2SO_4}=n_{H_2}=0,02\left(mol\right)\)
=> \(C\%_{H_2SO_4}=\dfrac{0,02.98}{200}.100=0,98\%\)
200ml = 0,2l
\(n_{HCl}=0,5.0,2=0,1\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,1 0,05
b) \(n_{Mg}=\dfrac{0,1.2}{1}=0,05\left(mol\right)\)
⇒ \(m_{Mg}=0,05.24=1,2\left(g\right)\)
c) \(n_{H2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
d) \(n_{MgCl2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(C_{M_{MgCl2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
Chúc bạn học tốt
`a)PTHH:`
`Zn + 2HCl -> ZnCl_2 + H_2`
`0,1` `0,2` `0,1` `0,1` `(mol)`
`n_[HCl]=0,2.1=0,2(mol)`
`=>m_[Zn]=0,1.65=6,5(g)`
`b)m_[dd HCl]=1,1.200=220(g)`
`=>C%_[ZnCl_2]=[0,1.136]/[6,5+220-0,1.2].100~~6%`
\(a,n_{HCl}=0,2.1=0,2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1<--0,2------>0,1------->0,1
\(\rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)
\(b,m_{ddHCl}=200.1,1=220\left(g\right)\)
\(\rightarrow m_{dd}=220+6,5-0,1.2=226,3\left(g\right)\\ \rightarrow C\%_{ZnCl_2}=\dfrac{0,1.136}{226,3}.100\%=6\%\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
0,3 0,3 0,3 0,3
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(m_{Fe}=0,3.56=16,8\left(g\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)
\(C_{M_{FeSO_4}}=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)
PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂
200ml = 0,2 lít.
a) Số mol H₂: nH₂ = 6,72 ÷ 22,4 = 0,3 mol
Theo PTHH => Số mol Fe: nFe = 0,3 mol
=> Khối lượng Fe: mFe = 16,8g
b) Số mol H₂SO₄: nH₂SO₄ = 0,3 mol
Nồng độ mol dd: CM = 0,3 ÷ 0,2 = 1,5M
a)
$Mg + 2HCl \to MgCl_2 + H_2$
Theo PTHH :
$n_{MgCl_2} = n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$m_{MgCl_2} = 0,25.95 = 23,75(gam)$
b)
$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{dd\ HCl} = \dfrac{0,5.36,5}{7,3\%} = 250(gam)$
c)
$2K + 2H_2O \to 2KOH + H_2$
$n_K = 2n_{H_2} = 0,5(mol)$
$m_K = 0,5.39 = 19,5(gam)$
Câu 1:
a) PTHH: \(2A+2xHCl\rightarrow2ACl_x+xH_2\uparrow\)
\(2B+2yHCl\rightarrow2BCl_y+yH_2\uparrow\)
b) Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\) \(\Rightarrow m_{H_2}=0,4\cdot2=0,8\left(g\right)\)
Theo các PTHH: \(n_{HCl}=2n_{H_2}=0,8\left(mol\right)\) \(\Rightarrow m_{HCl}=0,8\cdot36,5=29,2\left(g\right)\)
Bảo toàn khối lượng: \(a=m_{KL}=m_{muối}+m_{H_2}-m_{HCl}=38,6\left(g\right)\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
b, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\Rightarrow m_{ZnO}=21,1-13=8,1\left(g\right)\)
c, \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Zn}+2n_{ZnO}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
b) \(m_{Zn}=0,2.65=13\left(g\right)\)
→ \(m_{ZnO}=21,1-13=8,1\left(g\right)\)
\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
\(n_{HCl}=2.0,1+2.0,2=0,6\left(mol\right)\)
200ml = 0,2l
c) \(C_{MddH2SO4}=\dfrac{0,6}{0,2}=3\left(M\right)\)
Chúc bạn học tốt