Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right);n_{ZnCl_2}=\dfrac{27,2}{136}=0,2\left(mol\right)\)
PTHH(1): 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,2 0,3
PTHH(2): Zn + 2HCl → ZnCl2 + H2
Mol: 0,2 0,2 0,2
Ta có: \(n_{H_2\left(1\right)}=0,5-0,2=0,3\left(mol\right)\)
\(m_{hh}=0,2.27+0,2.65=18,4\left(g\right)\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
a, Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=1,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)
b, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
a)
\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ Zn + 2HCl \to ZnCl_2 + H_2\)
b)
Gọi : \(n_{H_2} = a(mol) \Rightarrow n_{HCl} = 2a\)
Bảo toàn khối lượng :
\(13,5 + 2a.36,5 = 66,75 + 2.a\\ \Rightarrow a = 0,75\\ \Rightarrow V = 0,75.22,4 = 16,8(lít)\)
a) Mg + 2 HCl -> MgCl2 + H2
2Al + 6 HCl -> 2 AlCl3 + 3 H2
Fe + 2 HCl -> FeCl2 + H2
Zn + 2 HCl -> ZnCl2 + H2
b) mY-mX=mCl
<=> mCl= 66,75-13,5=53,25(g)
=>nCl=53,25/35,5=1,5(mol)
=> nH2= nCl/2= 1,5/2=0,75(mol)
=>V=V(H2,đktc)=0,75.22,4=16,8(l)
pứ: Fe + 2HCl -> FeCl2 + H2
b. nFe = \(\dfrac{5,6}{56}\)= 0,1 mol
Từ pt suy ra được: nHCl = 2.nFe= 0,2 mol
=> mHCl = 0,2. 36,5 = 7,3 g
c. nH2 = nFe = 0,1 mol
=> VH2 = 0,1.22,4 = 2,24 (lít)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
+\(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
+\(nH_2=n_{Zn}=0,5\left(mol\right)\)
+\(n_{HCl}=2n_{Zn}=1\left(mol\right)\)
+\(V_{H2}=0,5.22,4=11,2\left(lit\right)\)
\(m_{HCl}=1.36,5=36,5\left(gam\right)\)
\(n_{Zn}=\dfrac{m}{M}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
\(Zn\) \(+\) \(2\)\(HCl\) → \(ZnCl_2\) \(+\) \(H_2\)
\(0,5\) \(mol\) → \(1\) \(mol\) → \(0,5\)\(mol\) → \(0,5\) \(mol\)
\(V_{H_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)
\(m_{HCl}=n.M=1.36,5=36,4\left(g\right)\)
Gọi nAl = nZn = a (mol)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a---------------------->1,5a
Zn + 2HCl --> ZnCl2 + H2
a--------------------->a
=> 1,5a + a = 0,6
=> a = 0,24 (mol)
=> mhh = 0,24.27 + 0,24.65 = 22,08 (g)
Gọi \(n_{Al}=n_{Zn}=a\left(mol\right)\)
PTHH:
2Al + 6HCl ---> 2AlCl3 + 3H2
a a
Zn + 2HCl ---> ZnCl2 + H2
a a
\(\rightarrow22,4\left(a+a\right)=13,44\\ \Leftrightarrow a=0,3\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}m_{Al}=0,3.27=8,1\left(g\right)\\m_{Zn}=0,3.65=19,5\left(g\right)\end{matrix}\right.\\ \rightarrow m_{hh}=8,1+19,5=27,6\left(g\right)\)
bạn xem lại đề đi
la 1,4 g H2
xin loi cau nhieu nha