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\(n_O=\dfrac{33,3-21,3}{16}=0,75\left(mol\right)\)
=> nH2O = 0,75 (mol)
Giả sử có V lít dd
=> \(\left\{{}\begin{matrix}n_{H_2SO_4}=V\left(mol\right)\\n_{HCl}=2V\left(mol\right)\end{matrix}\right.\)
Bảo toàn H: 2V + 2V = 0,75.2
=> V = 0,375 (lít) = 375 (ml)
$2Mg + O_2 \xrightarrow{t^o} 2MgO$
$2Cu + O_2 \xrightarrow{t^o} 2CuO$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$MgO + 2HCl \to MgCl_2 + H_2O$
$CuO + 2HCl \to CuCl_2 + H_2O$
$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$
Gọi $n_{MgO} = a(mol) ; n_{CuO} = b(mol) ; n_{Al_2O_3} = c(mol)$
Bảo toàn khối lượng : $m_{O_2} = 23,2 - 16,8 = 6,4(gam)$
$n_{O_2} = 0,2(mol)$
$\Rightarrow 0,5a + 0,5b + 1,5c = 0,2(1)$
Theo PTHH :
$n_{HCl} =2 n_{MgO} + 2n_{CuO} + 6n_{Al_2O_3} = 0,8(theo (1))$
Suy ra : $V_{dd\ HCl} = \dfrac{0,8}{2} = 0,4(lít)$
\(n_{HCl}=0,5a\left(mol\right)\)
PTHH:
2Al + 6HCl ---> 2AlCl3 + 3H2
Zn + 2HCl ---> ZnCl2 + H2
Mg + 2HCl ---> MgCl2 + H2
Fe + 2HCl ---> FeCl2 + H2
Theo các pthh: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,5a=0,25a\left(mol\right)\)
\(n_{H_2\left(pư\right)}=0,25a.80\%=0,2a\left(mol\right)\)
\(m_{giảm}=m_O=40-36,8=3,2\left(g\right)\)
Bảo toàn O: \(n_{H_2\left(pư\right)}=n_O=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(\rightarrow0,2a=0,1\Leftrightarrow a=2\)
\(m_O=22.3-14.3=8\left(g\right)\)
\(n_O=\dfrac{8}{16}=0.5\left(mol\right)\)
Bảo toàn nguyên tố O :
\(n_{H_2O}=n_O=0.5\left(mol\right)\)
Bảo toàn nguyên tố H :
\(n_{HCl}=2n_{H_2O}=0.5\cdot2=1\left(mol\right)\)
\(V_{dd_{HCl}}=\dfrac{1}{2}=0.5\left(l\right)\)
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\) (1)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (2)
\(2Cu+O_2\underrightarrow{t^o}2CuO\) (3)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\) (4)
Gọi: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\\n_{Cu}=z\left(mol\right)\end{matrix}\right.\) ⇒ 24x + 27y + 64z = 1,384 (1)
Ta có: \(n_{H_2}=\dfrac{0,3584}{22,4}=0,016\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=0,016\left(2\right)\)
\(n_{HCl\left(\left(1\right)+\left(2\right)\right)}=2n_{H_2}=0,032\left(mol\right)=n_{HCl\left(4\right)}\) \(n_{Cu}=n_{CuO}=\dfrac{1}{2}n_{HCl\left(4\right)}=0,016=z\left(3\right)\)
Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}x=0,012\left(mol\right)\\y=\dfrac{1}{375}\left(mol\right)\\z=0,016\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,012.24=0,288\left(g\right)\\m_{Al}=\dfrac{1}{375}.27=0,072\left(g\right)\\m_{Cu}=0,016.64=1,024\left(g\right)\end{matrix}\right.\)
b, \(C_{M_{HCl}}=\dfrac{0,032}{0,32}=0,1\left(M\right)\)
Gọi CT oxit sắt là FexOy
Gọi nCu=a(mol)
nH2=\(\dfrac{6,72}{22,4}\)=0,3(mol)
FexOy+yH2to→xFe+yH2O(1)
Fe+2HCl→FeCl2+H2(2)
Theo pthh(2)
nFe=nH2=0,3(mol)
Theo pthh(1)
nFexOy=\(\dfrac{0,3}{x}\)(mol)
Ta có: 64a+56.0,3=29,6
⇒a=0,2(mol)
⇒mCu=0,2.64=12,8(g)
⇒mFexOy=36−12,8=23,2(g)
=>MFexOy= \(\dfrac{\dfrac{23,2}{0,3}}{x}\)=\(\dfrac{232x}{3}\)
=>56x+16y=\(\dfrac{232x}{3}\)
=>\(\dfrac{64x}{3}=16y\)
->\(\dfrac{x}{y}=\dfrac{3}{4}\)
⇒CTHH:Fe3O4
Ta có :
%m Cu=\(\dfrac{12,8}{36}100\)=35,56%
=>%m Fe3O4=100%-35,56%=64,44%
tham khảo
Bảo toàn khối lượng:
m kim loại+ mO2= moxit
=> mO2= 3.33-2.13=1.2g
=> nO2= 1.2/32=0.0375mol
=>nO=0.075mol
mà cứ 1O + 2H+ = 1H2O
=> 0.075mol 0.15mol
vậy nH+ cần dùng là 0.15mol
mà CM=n / V => V= n / CM = 0.15 / 2 = 0.075l =75ml