Cho 5,4g Al tác dụng 100ml dd \(H_2SO_4\)
a.Viết pt. Tính\(V_{H_2}\)(ĐKTC)
b.Tính \(\)Cm\(H_2SO_4\), Khối lượng muối
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\(n_{Mg}=a\left(mol\right)\)
\(n_{Al}=b\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0.5\left(mol\right)\)
Giả sử : hỗn hợp chỉ có Mg
\(n_{Mg}=\dfrac{7.8}{24}=0.325\left(mol\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(0.325......0.325\)
\(n_{H_2SO_4}=0.325< 0.5\left(1\right)\)
Giả sử : hỗn hợp chỉ có Al.
\(n_{Al}=\dfrac{7.8}{27}=0.289\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.289....0.4335\)
\(n_{H_2SO_4}=0.4335\left(mol\right)< 0.5\left(2\right)\)
\(\left(1\right),\left(2\right):\)
Hỗn hợp tan hết , axit dư
\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(\left\{{}\begin{matrix}24a+27b=7.8\\a+1.5b=0.4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.2\end{matrix}\right.\)
\(\%Mg=\dfrac{0.1\cdot24}{7.8}\cdot100\%=30.77\%\)
\(\%Al=69.23\%\)
1. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
2. Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 9,2 (1)
Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(2\right)\)
Từ (1) và (2) ⇒ x = y = 0,1 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%\approx70,65\%\\\%m_{Al}\approx29,35\%\end{matrix}\right.\)
3. Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnSO_4}=0,1.160=16\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\end{matrix}\right.\)
1. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
2. Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 17,7 (1)
Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{66}{235}\\y=-\dfrac{29}{1410}\end{matrix}\right.\)
Tới đây thì ra số mol âm, bạn xem lại đề nhé.
a) Zn + H2SO4 --> ZnSO4 + H2
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
0,4-->0,4------->0,4---->0,4
=> \(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)
b) \(m_{ZnSO_4}=0,4.161=64,4\left(g\right)\)
c) \(\left\{{}\begin{matrix}m_{H_2}=0,4.2=0,8\left(g\right)\\V_{H_2}=0,4.22,4=8,96\left(l\right)\end{matrix}\right.\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\
pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,4 0,4 0,4
\(m_{H_2SO_4}=0,8.98=78,4\left(g\right)\\
m_{ZnSO_4}=136.0,4=54,4\left(g\right)\\
m_{H_2}=0,4.2=0,6\left(g\right)\\
V_{H_2}=0,4.22,4=8,96\left(l\right)\)
Đổi:400ml=0,4l
Gọi x;2y là số mol Fe,Al
Theo gt:\(m_{hhKL}\)=\(m_{Fe}+m_{Al}\)=56x+27y.2
=56x+54y=11(1)
Ta có PTHH:
Fe+\(H_2SO_4\)->\(FeSO_4\)+\(H_2\)(1)
x..........x................x.................(mol)
4Al+6\(H_2SO_4\)->2\(Al_2(SO_4)_3\)+3\(H_2\)(2)
2y...........3y...............y.........................(mol)
Ta có:\(C_{MddH_2SO_4}\)=1M
=>\(n_{H_2SO_4}\)=1.0,4=0,4mol
Theo PTHH(1);(2):
\(n_{H_2SO_4}\)=x+3y=0,4(2)
Từ (1);(2)=>\(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m_{Al}=54y=54.0,1=5,4\left(g\right)\\m_{Fe}=56x=56.0,1=5,6\left(g\right)\end{matrix}\right.\)
Theo PTHH(1);(2):\(n_{FeSO_4}\)=x=0,1(mol)
\(n_{Al_2\left(SO_4\right)_3}\)=y=0,1(mol)
Vậy \(C_{M\left(FeSO_4\right)}\)=0,1:0,4=0,25M
\(C_{MAl_2\left(SO_4\right)_3}\)=0,1:0,4=0,25M
\(n_{H_2SO_4}=0,4.1=0,4\left(mol\right)\)
Gọi x, y lần lượt là số mol của Al, Fe
Pt: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)
x \(\rightarrow\dfrac{3x}{2}\) \(\rightarrow0,1mol\)
Pt: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (2)
y \(\rightarrow y\) \(\rightarrow0,1mol\)
(1)(2) \(\Rightarrow\left\{{}\begin{matrix}1,5x+y=0,4\\27x+56y=11\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,4}=0,25M\)
\(C_{M_{FeSO_4}}=\dfrac{0,1}{0,4}=0,25M\)
pt: 2Al + 3H2SO4 => Al2(SO4)3 + 3H2
nAl = \(\dfrac{5,4}{27}=0,2mol\)
a) Theo pt: nH2 = \(\dfrac{3}{2}nAl=\dfrac{3}{2}.0,2=0,3mol\)
=> VH2 = 0,3.22,4 = 6,72 lít
b) Theo pt : nAl2(SO4)3 = \(\dfrac{1}{2}nAl=0,1mol\)
=> mAl2SO4 = 0,1.342 = 34,2 g
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{KOH}=0,2\cdot1=0,2\left(mol\right)\\n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) Axit còn dư 0,1 mol
\(\Rightarrow n_{H_2}=0,1\left(mol\right)\) \(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
a) Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
b) V = 100ml = 0,1l
\(C_{MH2SO4}=\dfrac{0,3}{0,1}=3\left(M\right)\)
\(m_{muối}=m_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)
Chúc bạn học tốt
a,2Al + 3H2SO4 → Al2(SO4)3 + 3H2
nAl = 5,4 : 27 = 0,2mol
nH\(_2\)=0,2.3:2 =0,3mol
VH\(_2\) = 0,3.22,4 =6,72 l
b. nH\(_2\)SO\(_4\) = 0,2.3:2=0,3mol
CM H\(_2\)SO\(_4\) = 0,3:0,1 =3M
nAl\(_2\)(SO\(_4\))\(_3\) = 0,2:2 =0,1mol
m\(Al_2 (SO_4 ) _3\) =0,1. 342 =34,2g